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Chemistry Calculations 9. The molar gas volume and Avogadro's Law 9. The molar gas volume and Avogadro's Law calculations (re-edit)

Doc Brown's Chemistry - GCSE, IGCSE, GCE (basic A level), O Level, US grade 10-12  online Chemical Calculations

9. The molar gas volume in calculations, application of Avogadro's Law to chemical reactions involving gases

Note on two acronyms and associated units:

STP or s.t.p. is a largely defunct abbreviation for 'standard temperature and pressure', 0oC/273K temperature and a pressure of 1 atm/101.3 kPa (101325 Pa). (In bygone times 760 mm of Hg of a 'mercury' barometer/manometer)

NTP or RTP is a frequently used abbreviation for 'normal' laboratory room temperature and pressure, taken as 20oC/293K temperature and pressure of 1 atm/101.3 kPa pressure (101325 Pa), however many standard pieces of chemical data are based on 25oC/298K, so take care on units! (temperature in K = oC + 273)

See also chemistry calculations Part 10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants to products) 9. The molar gas volume in calculations - moles, gas volumes and Avogadro's Law  It is better to know how to rearrange an equation than use a formula triangle

A summary of how to do basic molar gas volume calculations and rearrangement of the formula needed to solve problems

• Avogadro's Law states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules (irrespective of the nature of the gaseous particles)

• This means equal amounts of moles of gases occupy the same volume under the same conditions of temperature and pressure.

• So the volumes have equal moles of separate particles (molecules or individual atoms) in them.

• Therefore one mole of any gas (formula mass in g), at the same temperature and pressure occupies the same volume .

• This is 24 dm3 (24 litres) or 24000 cm3, at room temperature of 20oC/293K and normal pressure of 101.3 kPa/1 atmosphere (such conditions are often referred to as RTP/NTP).

• 20oC (293K) is often treated as room temperature, and the molar gas volume is 24.0 dm3 at normal pressure.

• 25oC (298K) is treated as a standard temperature, at which the molar volume of the gas is 24.4 dm3 at normal pressure.

• The molar volume for s.t.p is 22.4 dm3 (22.4 litres) at 0oC and 101 kPa1 atmosphere pressure.

• Historically, s.t.p unfortunately stands for standard temperature and pressure, but these days 25oC/298K is usually considered the standard temperature for data e.g. thermochemical and thermodynamic data at a higher academic level.

• Unfortunately, on the internet, both 20oC and 25oC are referred to as 'room temperature'.

• The Avogadro number is 6.02 x 1023 particles/mole of defined chemical 'species'.

• Therefore, e.g. at 20oC, 24.0 dm3 of a gas contains 6.03 x 1023 molecules.

• 6.03 x 1023 /  24.0 = 2.5 x 1022 = number of molecules in 1 dm3 of gas.

• 2.5 x 1022 / 1000 = 2.5 x 1019 = number of molecules in 1 cm3 of gas.

• Some handy (essential) relationships for substance Z below:

• It is much better to be able to rearrange an equation than use a triangle.

• moles Z = mass of Z gas (g) / atomic or formula mass of gas Z (g/mol)

• mass of Z in g = moles of Z x atomic or formula mass of Z

• atomic or formula mass of Z = mass of Z / moles of Z

• 1 mole = formula mass of Z  in g.

• gas volume of Z = moles of Z x volume of 1 mole

• rearranging this equation gives ...

• moles of Z = gas volume of Z / volume of 1 mole

• moles = V(dm3) / 24   (at RTP)

• The latter form of the equation can be used to calculate molecular mass from experimental data because

• moles = mass / molecular mass = gas volume / volume of 1 mole

• mass / molecular mass = gas volume / volume of 1 mole

• molecular mass = mass x volume of 1 mole/volume

• therefore at RTP: Mr = mass(g) x 24 / V(dm3)

• so, if you know the mass of a gas and its volume, you can work out moles of gas and then work out molecular mass.

• This has been done experimentally in the past, but these days, molecular mass is readily done very accurately in a mass spectrometer.

• Note (i): In the following examples, assume you are dealing with room temperature and pressure i.e. 25oC and 1 atmosphere pressure so the molar volume is 24dm3 or 24000cm3.

• Note (ii):

• Apart from solving the problems using the mole concept (method (a) below, and reading any equations involved in a 'molar way' ...

• It is also possible to solve them without using the mole concept (method (b) below). You still use the molar volume itself, but you think of it as the volume occupied by the formula mass of the gas in g and never think about moles!

• Methods of measuring how much gas is formed in a reaction

• (volume can be compared with theoretical prediction!)

• (a) • You can collect the gases in a calibrated gas syringe.

• You must make sure too much gas isn't produced and too fast!

• A gas syringe is more accurate than collecting the gas in an inverted measuring cylinder under water shown below, but its still only accurate to the nearest cm3.

• You can collect any gas by this method.

• (b) • The gas is collected in a measuring cylinder filled with water and inverted over a trough of water.

• You can get a more accurate result by using an inverted burette instead of a measuring cylinder.

• However, this method is no good if the gas is soluble in water!

• Burettes are calibrated in 0.10 cm3 intervals. measuring cylinders to the nearest cm3 or worse!

• In both methods the reaction is carried out in conical flask fitted with a sealing rubber bung, but a tube enabling the gas evolved to be collected in some suitable container.

• (c) • A third method is to measure the mass loss due to gas loss by carrying out the reaction in a flask set up on an accurate one-pan electronic balance.

• You need to put a cotton wool plug in the neck of the conical flask in case you lose any of the solution in a spray as the gas bubbles up - effervescence can produce an aerosol.

• This method can be used for any reaction that produces a gas, but the gas is released into the laboratory, ok if its harmless.

• It is potentially the most accurate method, BUT, the mass loss may be quite small especially hydrogen [Mr(H2) = 2], better for the 'heavier' gas carbon dioxide [MrCO2) = 44]

• There is also a possible error if the gas dissolves in water, oxygen has a very low solubility, but carbon dioxide is much more soluble and will reduce the measured mass of gas evolved and measured.

• In these gas volume calculations assume the molar gas volume is 24 dm3 (24 000 cm3) at 'normal' temperature and pressure.

• Molar gas volume calculation Example 9.1

• (a) What is the volume of 3.5 g of hydrogen (H2)? [Ar(H) = 1]

• common thinking: hydrogen exists as H2 molecules, so Mr(H2) = 2, so 1 mole or formula mass in g = 2 g

• method (i)

• so moles of hydrogen  = 3.5/2 = 1.75 mol H2

• so volume H2 = mol H2 x molar volume = 1.75 x 24 = 42 dm3 (or 42000 cm3)

• -

• method (ii):

• 2 g occupies 24 dm3, so scaling up for the volume of hydrogen ...

• 3.5 g will have a volume of 3.5/2  x 24 = 42 dm3 (or 42000 cm3)

• -

• (b) What is the volume of 11.6 g of butane gas at RTP?

• The formula of butane is C4H10, atomic masses are C = 12 and H = 1

• Formula mass of butane = (4 x 12) + (10 x 1) = 58

• moles butane = mass butane / formula mass = 11.6/58 = 0.20 mol

• volume = moles x molar volume

• volume butane = 0.20 x 24 = 4.8 dm3

• -

• (c) What mass of chlorine (Cl2) does 6.0 dm3 of the gas contain at rtp?

• Atomic mass of Cl = 35.5, molecular mass of chlorine = 2 x 35.5 = 71

• 1 mol gas = 24 dm3, so mol Cl2 = 6/24 = 0.25 mol

• mass = moles x molecular mass = 0.25 x 71 = 17.75 g (4sf, 2dp)

• -

• (d) What mass of methane (CH4) is contained in a 10 dm3 cylinder at three times normal atmospheric pressure at room temperature.

• Atomic masses: C = 12,  H = 1, molar gas volume = 24 dm3.

• Formula mass of methane = 16 + (4 x 1) = 16

• At normal pressure mol of CH4 would be 10/24 = 0.4166

• However, in 3x normal pressure there will be 3x as many methane particles in the same volume.

• So the actual number of moles at this higher pressure will be 3 x 0.4166 = 12.5 mol

• mass CH4 = mol CH4 x formula mass of CH4 = 12.5 x 16 = 200 g (or 0.20 kg)

• -

• From now on the calculations involve interpreting an equation in terms of moles e.g.

• (e) If 10 g of calcium carbonate (limestone) is thermally decomposed what volume of carbon dioxide is formed at room temperature and pressure?

• The equation is: CaCO3(s)  ===>  CaO(s)  +  CO2(g)

• Atomic masses: Ca = 40,  C = 12,  O =16. Formula mass CaCO3 = 40 + 12 + (3 x 16) = 100

• 1 mol of calcium carbonate gives 1 mol of carbon dioxide.

• Therefore mol CaCO3 = mol CO2 = 10/100 = 0.10 mol

• Therefore volume CO2 = mol CO2 x molar gas volume

• Volume of CO2 = 0.10 x 24.0 = 2.4 dm3 (or 2.4 x 1000 = 2400 cm3)

• -

• AND from now on, you have to the questions - ANSWERS at bottom of page

• AND remember it is better to be able to rearrange a formula than use a triangle.

• AND do not assume all atomic masses quoted are needed to solve the problem.

• Molar gas volume calculation Example 9.2 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• Given the equation

• MgCO3(s) + H2SO4(aq) ==> MgSO4(aq) + H2O(l) + CO2(g)

• This equation is read as 1 mole of magnesium carbonate reacts with 1 mole of sulfuric acid to form 1 mole of magnesium sulfate, 1 mole of water and 1 mole of carbon dioxide gas

• What mass of magnesium carbonate is needed to make 6 dm3 of carbon dioxide?

• [Ar's: Mg = 24, C = 12, O = 16, H =1 and S = 32]

• The important mole ratio is 1 MgCO3 ==> 1 CO2

• Fully worked out ANSWERS at the bottom of the page

• -

• Molar gas volume calculation Example 9.3 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• 6g of a hydrocarbon gas had a volume of 4.8 dm3, calculate its molecular mass.

• Fully worked out ANSWERS at the bottom of the page

-

• Molar gas volume calculation Example 9.4 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• Given the equation ... (and Ar's Ca = 40, H = 1, Cl = 35.5)

• Ca(s) + 2HCl(aq) ==> CaCl2(aq) + H2(g)

• The equation is read as 1 mole of calcium atoms reacts with 2 moles of hydrochloric acid to form 1 mole of calcium chloride salt and 1 mole of hydrogen gas molecules (NOT atoms).

• What volume of hydrogen is formed when ...

• (i) 3g of calcium is dissolved in excess hydrochloric acid?

• The important mole ratio is 1 Ca ==> 1 H2

• (ii) 0.25 moles of hydrochloric acid reacts with calcium?

• The important mole ratio is 1 Ca + 2HCl

• Molar gas volume calculation Example 9.5 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• Given the equation ... (and Ar's Mg = 24, H = 1, Cl = 35.5)

• Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g)

• The equation is read as 1 mole of magnesium reacts with 2 moles of hydrochloric acid to form 1 mole of magnesium chloride salt and 1 mole of hydrogen gas molecules (NOT atoms).

• How much magnesium is needed to make 300 cm3 of hydrogen gas?

• Fully worked out ANSWERS at the bottom of the page

• -

• Molar gas volume calculation Example 9.6 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• A small teaspoon of sodium hydrogencarbonate (baking soda) weighs 4.2g.

• Calculate the moles, mass and volume of carbon dioxide formed when it is thermally decomposed in the oven.

• Assume room temperature for the purpose of the calculation.

• 2NaHCO3(s) ==> Na2CO3(s) + H2O(g) + CO2(g)

• atomic masses: Na = 23,  H = 1,  C = 12,  O = 16

• Fully worked out ANSWERS at the bottom of the page

• -

• Molar gas volume calculation Example 9.7 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• What volume of carbon dioxide is formed at RTP when 5g of carbon is completely burned?

• C(s) + O2(g) ==> CO2(g)

• atomic mass of carbon = 12  O = 16

• Fully worked out ANSWERS at the bottom of the page

• -

• Molar gas volume calculation Example 9.8 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• What volume of carbon dioxide gas is formed at RTP if 1Kg of propane gas fuel is burned?

• C3H8(g) + 5O2(g)  ==> 3CO2(g) + 4H2O(l)

• atomic masses: C = 12, O = 16  H =1

• Fully worked out ANSWERS at the bottom of the page

• -

• Molar gas volume calculation Example 9.9 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• 3.27g of metal dissolved in dilute sulfuric acid to form 1.2 dm3 of hydrogen gas at RTP.

• It was believed that the equation for the reaction is:

• M  +  H2SO4  ===>  MSO4  +  H2

• 1 mole of gas at RTP occupies a volume of 24 dm3

• Calculate the atomic mass of the metal.

• Fully worked out ANSWERS at the bottom of the page

• -

• Molar gas volume calculation Example 9.10

• Sodium hydrogencarbonate reacts with acids to form a salt, water and carbon dioxide e.g.

• sodium hydrogencarbonate  +  sulfuric acid  ===>  sodium sulfate  +  water  + carbon dioxide

• The balanced symbol equation:

• 2NaHCO3(s)  +  H2SO4(aq)  ===>  Na2SO4(aq)  +  2H2O(l)  +  2CO2(g)

• Atomic masses: Na =23,   H = 1,   S = 32,   O = 16

• (a) Calculate the formula masses of sodium hydrogencarbonate, sodium sulfate and carbon dioxide.

• -

• (b) What is the maximum volume of carbon dioxide that can be collected if 10.0 g of sodium hydrogencarbonate is dissolved in excess sulfuric acid?

• Fully worked out ANSWERS at the bottom of the page

• -

• (c) What mass of sodium sulfate is made assuming a 100% yield (hardly so in practice!)

• Fully worked out ANSWERS at the bottom of the page

• -

• Molar gas volume calculation Example 9.11

• Atomic masses: Na = 23,  H = 1,  O = 16  and the volume of 1 mole of gas is 24.0 dm3.

• Sodium reacts with excess water to form sodium hydroxide and hydrogen gas.

• If 4.6 g of sodium reacts with water, what volume of hydrogen is formed at room temperature and pressure?

• The equation is: 2Na(s)  +  2H2O(l)  ===>  2NaOH(aq)  +  H2(g)

• Fully worked out ANSWERS at the bottom of the page

Learning objectives for problem solving using the molar gas volume

Know that the volume for 1 mole of gas varies with temperature and pressure.

Know the different units of temperature and pressure and be able to convert from one to the other in the context of calculations of moles and volumes of gases.

Know how to solve problems and do calculations based on the volume of 1 mole of gas (molar volume, usually taken as 24 dm3 at GCSE level).

Know at RTP moles of gas = volume of gas in dm3 / 24 and be able to rearrange this moles of gas' equation.

Know that Avogadro's Law states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules.

Be able to describe and explain how to measure the volume of a gas formed in a chemical reaction e.g. collection in gas syringe or recording loss of mass from a system as a gas is produced and escapes into the atmosphere.

Know how to read equations in terms of (molar) mole ratios e.g. to deduce volumes of gases formed from a given mass of reactant - spot the crucial mole ratio in the equation.

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Worked out ANSWERS to the moles and gas volume questions

• Molar gas volume calculation Example 9.2 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• Given the equation

• MgCO3(s) + H2SO4(aq) ==> MgSO4(aq) + H2O(l) + CO2(g)

• This equation is read as 1 mole of magnesium carbonate reacts with 1 mole of sulfuric acid to form 1 mole of magnesium sulfate, 1 mole of water and 1 mole of carbon dioxide gas

• What mass of magnesium carbonate is needed to make 6 dm3 of carbon dioxide?

• [Ar's: Mg = 24, C = 12, O = 16, H =1 and S = 32]

• The important mole ratio is 1 MgCO3 ==> 1 CO2

• method (a):

• since 1 mole = 24 dm3, 6 dm3 is equal to 6/24 = 0.25 mol of gas

• From the equation, 1 mole of MgCO3 produces 1 mole of CO2, which occupies a volume of 24 dm3.

• so 0.25 moles of MgCO3 is needed to make 0.25 mol of CO2

• formula mass of MgCO3 = 24 + 12 + (3 x 16) = 84,

• so required mass of MgCO3 = mol x formula mass = 0.25 x 84 = 21g

• -

• method (b):

• converting the equation into the required reacting masses ..

• formula masses: MgCO3 = 84 (from above), CO2 = 12 + (2 x 16) = 44

• MgCO3 : CO2 equation ratio is 1 : 1

• so 84g of MgCO3 will form 44g of CO2

• 44g of CO2 will occupy 24dm3

• so scaling down, 6 dm3 of CO2 will have a mass of 44 x 8/24 = 11g

• if 84g MgCO3 ==> 44g of CO2, then ...

• 21g MgCO3 ==> 11g of CO2 by solving the ratio, scaling down by factor of 4

• -

• Molar gas volume calculation Example 9.3 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

•  6g of a hydrocarbon gas had a volume of 4.8 dm3, calculate its molecular mass.

• method (a):

• 1 mole = 24 dm3, so moles of gas = 4.8/24 = 0.2 mol

• molecular mass = mass in g / moles of gas

• Mr = 6 / 0.2 = 30

• i.e. if 6g = 0.2 mol, 1 mol must be equal to 30 g by scaling up

• -

• method (b):

• 6g occupies a volume of 4.8 dm3

• the formula mass in g occupies 24 dm3

• so scaling up the 6g in 4.8 dm3

• there will be 6 x 24/4.8 = 30 g in 24 dm3

• so the molecular or formula mass = 30

• -

• Molar gas volume calculation Example 9.4 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• Given the equation ... (and Ar's Ca = 40, H = 1, Cl = 35.5)

• Ca(s) + 2HCl(aq) ==> CaCl2(aq) + H2(g)

• The equation is read as 1 mole of calcium atoms reacts with 2 moles of hydrochloric acid to form 1 mole of calcium chloride salt and 1 mole of hydrogen gas molecules (NOT atoms).

• What volume of hydrogen is formed when ...

• (i) 3g of calcium is dissolved in excess hydrochloric acid?

• The important mole ratio is 1 Ca ==> 1 H2

• (ii) 0.25 moles of hydrochloric acid reacts with calcium?

• The important mole ratio is 1 Ca + 2HCl

• (i) method (a):

• 3g Ca = 3/40 = 0.075 mol Ca

• from 1 : 1 ratio in equation, 1 mol Ca produces 1 mol H2

• so 0.075 mol Ca produces 0.075 mol H2

• so volume H2 = 0.075 x 24 = 1.8 dm3 (or 1800 cm3)

• -

• (i) method (b):

• from equation 1 Ca ==> 1 H2 means 40g ==> 2g

• so scaling down: 3g Ca will produce 2 x 3/40 = 0.15g H2

• 2g H2 has a volume 24 dm3, so scaling down ...

• 0.15g H2 has a volume of (0.15/2) x 24 = 1.8 dm3 (or 1800 cm3)

• -

• (ii) method (a) only:

• from equation: 2 moles HCl ==> 1 mole H2 (mole ratio 2:1)

• so 0.25 mol HCl ==> 0.125 mol H2, volume 1 mole gas = 24 dm3

• so volume H2 = 0.125 x 24 = 3 dm3

• -

• Molar gas volume calculation Example 9.5 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• Given the equation ... (and Ar's Mg = 24, H = 1, Cl = 35.5)

• Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g)

• The equation is read as 1 mole of magnesium reacts with 2 moles of hydrochloric acid to form 1 mole of magnesium chloride salt and 1 mole of hydrogen gas molecules (NOT atoms).

• How much magnesium is needed to make 300 cm3 of hydrogen gas?

• method (a)

• The important mole ratio is 1 Mg ==> 1 H2

• 300 cm3 = 300/24000 = 0.0125 mol H2 (since 1 mol of any gas = 24000 cm3)

• from the equation 1 mole Mg ==> 1 mole H2

• so 0.0125 mole Mg needed to make 0.0125 mol H2

• so mass of Mg = mole Mg x Ar(Mg)

• so mass Mg needed = 0.0125 x 24 = 0.30 g

• -

• method (b)

• reaction ratio in equation is 1 Mg ==> 1 H2,

• so reacting mass ratio is 24g Mg ==> 2g H2,

• 2g H2 has a volume of 24000 cm3 (volume of formula mass in g)

• so scaling down: mass Mg needed = 24 x (300/24000) = 0.30 g

• -

• Molar gas volume calculation Example 9.6 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• A small teaspoon of sodium hydrogencarbonate (baking soda) weighs 4.2g.

• Calculate the moles, mass and volume of carbon dioxide formed when it is thermally decomposed in the oven.

• Assume room temperature for the purpose of the calculation.

• 2NaHCO3(s) ==> Na2CO3(s) + H2O(g) + CO2(g)

• This equation is read as 2 moles of sodium hydrogencarbonate decomposes to give 1 mole of sodium carbonate, 1 mole of water and 1 mole of carbon dioxide gas.

• The important mole ratio is 2 NaHCO3 ==> 1 CO2

• atomic masses: Na = 23,  H = 1,  C = 12,  O = 16

• Formula mass of NaHCO3 is 23+1+12+(3x16) = 84 = 84g/mole

• Formula mass of CO2 = 12+(2x16) = 44 = 44g/mole (not needed by this method)

• and a molar gas volume of 24000 cm3 at RTP (definitely needed for this method)

• In the equation 2 moles of NaHCO3 give 1 mole of CO2 (2:1 mole ratio in equation)

• Moles NaHCO3 = 4.2/84 = 0.05 moles ==> 0.05/2 = 0.025 mol CO2 on decomposition.

• Mass = moles x formula mass, so mass CO2 = 0.025 x 44 = 1.1g CO2

• Volume = moles x molar volume = 0.025 x 24000 = 600 cm3 of CO2

• -

• Molar gas volume calculation Example 9.7 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• What volume of carbon dioxide is formed at RTP when 5g of carbon is completely burned?

• C(s) + O2(g) ==> CO2(g)

• 1 mole carbon reacts with 1 mole of oxygen molecules (NOT atoms) gives 1 mole of carbon dioxide

• atomic mass of carbon = 12 O =16

• The important mole ratio is 1 C ==> 1 CO2

• moles = mass / atomic mass, moles carbon = moles carbon dioxide = 5/12 = 0.417 mol

• 1 mole of gas at RTP occupies 24 dm3

• so 0.417 mol occupies a volume of 0.417 x 24 = 10.0 dm3

• -

• Molar gas volume calculation Example 9.8 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• What volume of carbon dioxide gas is formed at RTP if 1Kg of propane gas fuel is burned?

• C3H8(g) + 5O2(g)  ==> 3CO2(g) + 4H2O(l)

• 1 mole of propane gas reacts with 5 moles of oxygen to gives 3 moles of carbon dioxide gas and 4 moles of water on complete combustion

• The important mole ratio is 1 C3H8 ==> 3 CO2

• 1 kg = 1000g, atomic masses: C = 12, H =1, O =16

• Mr(propane) = (3 x 12) + (8 x 1) = 44

• moles = mass in g / molecular mass, therefore moles propane = 1000/44 = 22.73 mol

• from equation molar ratio: moles carbon dioxide = 3 x moles of propane

• mol propane = 3 x 22.73 = 68.18 mol

• 1 mole of gas at RTP occupies a volume of 24 dm3

• so 68.18 mol of gas occupies a volume of 68.18 x 24 = 1636 dm3

• -

• Molar gas volume calculation Example 9.9 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• 3.27g of metal dissolved in dilute sulfuric acid to form 1.2 dm3 of hydrogen gas at RTP.

• It was believed that the equation for the reaction is:

• M  +  H2SO4  ===>  MSO4  +  H2

• 1 mole of gas at RTP occupies a volume of 24 dm3

• Calculate the atomic mass of the metal.

• moles of hydrogen = 1.2 / 24 = 0.05

• According to the equation 1 mole of metal produces 1 mole of hydrogen.

• Therefore moles of metal M also equals 0.05

• rearranging: moles = mass / atomic mass

• gives atomic mass = mass / moles

• therefore atomic mass of metal M = 3.27 / 0/05 = 65.4

• -

• Molar gas volume calculation Example 9.10 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• Sodium hydrogencarbonate reacts with acids to form a salt, water and carbon dioxide e.g.

• sodium hydrogencarbonate  +  sulfuric acid  ===>  sodium sulfate  +  water  + carbon dioxide

• The balanced symbol equation:

• 2NaHCO3(s)  +  H2SO4(aq)  ===>  Na2SO4(aq)  +  2H2O(l)  +  2CO2(g)

• Atomic masses: Na =23,   H = 1,   S = 32,   O = 16

• (a) Calculate the formula masses of sodium hydrogencarbonate, sodium sulfate and carbon dioxide.

• Mr(NaHCO3)  =  23 + 1 + 12 + 3x16 = 84

• Mr(Na2SO4) = 2x23 + 32 + 4x16 = 142

• Mr(CO2) =  12 + 2x16 = 44

• -

•  (b) What is the maximum volume of carbon dioxide that can be collected if 10.0 g of sodium hydrogencarbonate is dissolved in excess sulfuric acid?

• From the equation 2 mol of the carbonate gives 2 mol of the gas.

• Therefore mol NaHCO3 = mol CO2 = 10.0/84 = 0.119 mol

• volume of CO2 = mol CO2 x molar gas volume

• volume CO2 = 0.119 x 24 = 2.86 dm3 (3sf, 2.86 x 1000 = 2860 cm3)

• -

• (c) What mass of sodium sulfate is made assuming a 100% yield (hardly so in practice!)

• From the equation 2 mol of NaHCO3 produces 1 mol of Na2SO4

• So mol of Na2SO4 = mol NaHCO3/2 = 0.119/2 = 0.0595

• mass Na2SO4 = mol Na2SO4 x formula mass of Na2SO4 = 0.0595 x 142 = 8.45 g (3 sf)

• -

• Molar gas volume calculation Example 9.11 (At RTP 1 mol gas = 24 dm3 or 24000 cm3)

• Atomic masses: Na = 23,  H = 1,  O = 16  and the volume of 1 mole of gas is 24.0 dm3.

• Sodium reacts with excess water to form sodium hydroxide and hydrogen gas.

• If 4.6 g of sodium reacts with water, what volume of hydrogen is formed at room temperature and pressure?

• The equation is: 2Na(s)  +  2H2O(l)  ===>  2NaOH(aq)  +  H2(g)

• From the equation 2 moles of sodium produce 1 mole of hydrogen.

• mol Na = 4.6/23 = 0.20

• Therefore mol hydrogen produced = 0.20/2 = 0.10

• So volume of hydrogen formed = 0.1 x 24 = 2.4 dm3

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