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'.
-
Reminder about the Avogadro Number:
-
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
-
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)
-
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?
-
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
...
-
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)
-
Molar gas volume calculation
Example 9.8 (At RTP 1 mol gas = 24
dm3 or 24000 cm3)
-
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:
-
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:
-
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?
-
(c) What mass of sodium sulfate is made
assuming a 100% yield (hardly so in practice!)
-
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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QUIZ
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Notes on
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Part 10.
Reacting gas volume
ratios, Avogadro's Law
& Gay-Lussac's Law
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OTHER CALCULATION PAGES
1.
What is relative atomic mass, relative isotopic mass, calculating relative atomic mass
2.
Calculating relative
formula/molecular mass of a compound or element molecule
3.
Law of Conservation of Mass and simple reacting mass calculations
4. Composition by percentage mass of elements
in a compound
5. Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
6a. Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination (see calculations section 14.)
6b.
Reacting masses, concentration of solution and volumetric titration calculations
(NOT using moles)
7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)
8. Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
9.
Moles and the molar volume of a gas, Avogadro's Law
10. Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law (ratio of gaseous
reactants to products)
11. Molarity, volumes and solution
concentrations (and diagrams of apparatus)
(this page)
12.
How to do acid-alkali
titration calculations, diagrams of apparatus, details of procedures
13. Electrolysis products calculations (negative cathode and positive anode products)
14. Other calculations
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
14.1
% purity of a product 14.2a
% reaction yield
14.2b
atom economy 14.3
dilution of solutions
14.4
water of crystallisation
calculation
14.5
how
much of a reactant is needed? limiting reactant calculations
Energy transfers in physical/chemical changes,
exothermic/endothermic reactions
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
Radioactivity and half-life calculations including
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calculations, how to measure the volume of gas given off in a chemical
reaction and calculation of moles of gas, working out moles of gas formed
given the chemical equation using Avogadro's law and molar volume practice
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level chemistry question exercise on calculating volumes of gases using
molar gas volume
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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?
-
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)
-
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) 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
-
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)
-
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)
-
Molar gas volume calculation
Example 9.8 (At RTP 1 mol gas = 24
dm3 or 24000 cm3)
-
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:
-
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:
-
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?
-
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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Phil Brown 2000+.
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Copying of website material is NOT
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are unofficial.
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