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Calculating Quantities of Reactants Needed

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

study examples carefully14. Other GCSE chemical calculations - CALCULATING THE QUANTITIES OF REACTANTS NEEDED for a chemical preparation reaction

and LIMITING REACTANT QUANTITIESstudy examples carefully

soluble salt preparation from insoluble base-acid neutralisationQuantitative chemistry calculations This page explains how to calculate the quantities of reactants needed to prepare a given amount of product. Help in how to theoretically calculate quantities of reactants needed and products formed. Mass of reactants needed for a chemical preparation - fully worked out example calculations. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses.  These revision notes and practice questions on how to do chemical calculations and worked examples what amounts of reactants are needed should prove useful for the new AQA, Edexcel and OCR GCSE (91) chemistry science courses.

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study examples carefully14. Other GCSE chemical calculationsstudy examples carefully

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


14.5 (a) Calculation of quantities required for a chemical reaction

and a brief mention of % yield and atom economy

(a) In a chemical preparation reaction involving two reactants, it is common to use an excess of one of the reactants to ensure that all of the other reactant is used but this still requires you to be able to calculate the quantities of chemicals needed.

(b) The reactant chemical that is completely used up is called the limiting reactant because it limits the amount of products after it has all been used up. Therefore you need to be able to explain the effect of a limiting quantity of a reactant on the maximum amount of products it is theoretically possible to obtain in terms of amounts in moles or masses in grams.

BUT, don't forget, you still never get in reality 100% yield.


Calculations of quantities of chemicals required to do a preparation
 

  • Example 1. How much iron and sulfur do you need to heat together to make 20.0 g of iron sulfide

    • Atomic masses: Fe = 56 and S = 32

    • Balanced equation: Fe + S ==> FeS

    • This is essentially a reacting mass calculation
      Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product comments
      Fe + S ==> FeS You need the balanced chemical equation and in this case its very simple.
      56g + 32g ==> 88g basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses
      56/88 = 0.6364g + 32/88 = 0.3636 ==> 88g/88 = 1.00g divide by 88 to scale down to 1g of FeS product
      20 x 0.6364 = 12.7g + 20 x 0.3636 = 7.3g ==> 20 x 100 = 20.0g then scale by a factor of 20 (for 20g of FeS)
      Therefore to make 20g of iron sulfide you need 12.7g of iron and 7.3g of sulfur.
    • -

  • Example 2. How much iron do you need to make 100.00g of iron(III) chloride by passing excess chlorine gas over heated iron filings?

    • Atomic masses: Fe = 56 and Cl = 35.5, formula mass of FeCl3 = 162.5 (56 + 3x35.5)

    • Balanced equation: 2Fe(s) + 3Cl2(g) ==> 2FeCl3(s)

    • This is essentially a reacting mass calculation
      Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product comments
      2Fe ==> 2FeCl3 You need the balanced chemical equation BUT only these bits of the equation are needed to solve the problem, you can ignore the rest of the equation
      2 x 56 = 112g ==> 2 x 162.5 = 325g basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses
      112/325 = 0.3446g ==> 325/325 = 1.00g divide by 325 to scale down to 1g of FeCl3 product
      100 x 0.3446 = 34.46g ==> 100 x 1 = 100.00g then scale by a factor of 100 to make 100g of FeCl3
      Therefore to make 100g of iron(III) chloride you need 34.46g of iron
    • -

  • Example 3 A more complex example based on a salt preparation

  • Suppose you want to make 50g of the blue hydrated copper(II) sulphate crystals

    • All of section 14.5 is based on this quantity of 50g of the familiar blue crystals.

    • Your reactants are dilute sulphuric acid and the black solid copper(II) oxide.

    • You can use copper(II) carbonate, but this is not a pure simple compound and the predictive nature of the calculations will not be as good.

    • copper(II) oxide + sulphuric acid ==> copper(II) sulphate + water

    • (i) CuO(s) + H2SO4(aq) ==> CuSO4(aq) + H2O(l)

    • on crystallisation you get the blue hydrated crystals of formula CuSO4.5H2O

    • so strictly speaking, after evaporation-crystallisation the overall equation is

    • (ii) CuO(s) + H2SO4(aq) + 4H2O(l) ==> CuSO4.5H2O(s)

    •    (formula of the blue crystals)
    • How much copper(II) oxide is needed?

    • A 'non-moles' calculation first of all, involving a reacting mass calculation.

    • The crucial change overall is CuO ==> CuSO4.5H2O

    • (Note: In reacting mass calculations you can often ignore other reactant/product masses)

    • Atomic masses: Cu = 64, S = 32, H = 1, O = 16

    • Formula masses are for: CuO = 64 + 16 = 80, CuSO4 = 64 + 32 + (4 x 16) = 160

    • and CuSO4.5H2O = 64 + 32 + (4 x 16) + [7 x (1+1+16)] = 250

    • The crucial reacting mass ratio is: 80 ==> 250 since formula ratio is 1:1 in the equation.

    • Therefore, theoretically, to make 50g of the crystals (1/5th of 250),

    • you need 1/5th of 80g of copper(II) oxide,

    • and 80/5 = 16g of copper(II) oxide is required.

    • However, in reality, things are not so simple because the method involves adding excess copper(II) oxide to the dilute sulphuric acid. (see salt preparation method (b) and see the note on 'limiting reactant', the last section of this  page.

    • So in practice you would need to use more of the CuO to get anything like 50g of the salt crystals.

    • There is another way to calculate the quantities required based on the acid.

    • How much dilute sulphuric acid (of concentration 1 mol dm-3) is required?

      • Mol = mass in g / formula mass,

      • so moles of CuSO4.5H2O required = 50/250 = 0.2 mol (see basics of moles)

      • Therefore 0.2 mol of H2SO4 is required (1/5th mol), since the mole ratio CuO : H2SO4 is 1 : 1 in the equation.

      • 1dm3 (1000 cm3) of a 1 mol dm-3 solution of contains 1 mole (by definition - see molarity page)

      • Therefore 1/5th of 1dm3 is required to provide 1/5th of mole of the sulfuric acid.

      • so 200 cm3 of 1 mol dm-3 dilute sulphuric acid is required,

      • or 100 cm3 of 2M dilute sulphuric acid. (2M is old fashioned notation for 2 mol dm-3 still seen on many laboratory bottles!)

      • You would then add copper(II) oxide in small amounts until no more dissolves in the warm-hot acid and the excess black powder is filtered off. There is no need to weigh out an exact amount of copper oxide.

      • If you want just 25g of copper sulfate crystals you would use 100 cm3 of 1 molar sulfuric acid, or 50cm3 of 2 molar sulfuric acid.

        • BUT REMEMBER

        • (i) in practice, you will NOT get a 100% yield, see calculation below.

        • (ii) It would be normal to use excess of the copper oxide, because it is easy to separate by filtration the unreacted oxide to leave a neutral solution of the salt.

    • Suppose after carrying out the preparation you finally crystallise 34g of pure the blue crystals of CuSO4.5H2O after weighing the dry product.

      • What is the 'atom economy' of the preparation? (you need to refer to equations (i)/(ii)  at the start of section 14.5

        • Atom economy = useful theoretical products x 100/mass of all reactants

        • based on equation (i) Atom economy = mass CuSO4 x 100 / (mass CuO + H2SO4)

        • = 160 x 100 / (80 + 98) = 16000/178 = 89.9%

        • based on equation (ii) the atom economy is 100% if you include water as a 'reactant', can you see why?

      • What is the % yield? i.e. comparing what you actually get with the maximum possible, i.e. a 'reality check'!

        • % yield = mass of product obtained x 100 / theoretical mass from the equation

        • % yield = 34 x 100 / 50 =  68%

    • More examples of % yield and atom economy calculations in section 6.

    • -


14.5 (b) Calculating limiting quantities of a reactant

Logically deducing from given quantities of reactants and the chemical equation, which reactant is the limiting reactant?

In a preparation chemical reaction involving two reactants, it is quite common to use an excess of one of the reactants to ensure that all of the other reactant is used up.

The reactant that is completely used up is called the limiting reactant because it limits the amount of products that can be formed i.e. when it is all used up in the reaction.

You need to be able to explain the effect of a limiting quantity of a reactant on the amount of products it is possible to make e.g. in terms of amount of product in moles or masses in grams.

 

Example 1. of a limiting reactant - making iron sulfide

If 40 g of sulfur is heated with 60g of iron filings to make iron sulfide, which is the limiting reactant? and what is the maximum mass of iron sulfide that can be made?

The reaction is: Fe + S ===>  FeS;  atomic masses: Fe = 56;  S = 32

 

(a) Calculating the limiting reactant

moles = mass in g / formula mass

moles S = 40 / 32 = 1.25 mol S

moles Fe = 56 / 60 = 0.933 mol Fe

Since one mole of iron reacts with one mole of sulfur, there is insufficient iron to react with all of the sulfur used in the experiment, therefore the iron is the limiting reactant.

 

(b) Calculating the maximum yield from the limiting reactant.

This part of the calculation MUST be based on the limiting reactant (0.933 mol Fe)

Formula mass iron sulfide FeS = 56 + 32 = 88

From the reaction equation, reading in molar quantities, one mole of iron forms one mole of iron sulfide.

mass FeS = mol FeS x f. mass FeS

mass FeS = 0.933 x 88 = 82.1 g

 

Example 2. of a limiting reactant - making copper sulfate from copper oxide and sulfuric acid

If 10 g of copper oxide (CuO) is added to 50 cm3 of sulfuric acid of concentration 2.0 mol/dm3, deduce which is the limiting reactant and the maximum amount of copper sulfate (as CuSO4) you could theoretically make?

The reaction is: CuO(s)  +  H2SO4(aq)  ===> CuSO4(aq)  + H2O(l)

Atomic masses: Cu = 64; O = 16; H = 1; S =32 (but you might not need all of these!)

 

(a) Deducing the limiting reactant

Formula mass copper sulfate = 64 + 16 = 80

moles = mass in g / formula mass

mol CuO = 10 / 80 = 0.125 mol

mol H2SO4 = molarity v volume (in dm3)    (dm3 = cm3 / 1000)

mol H2SO4 = 2.0 x 50 / 1000 = 0.10 mol

From the equation, reading in molar quantities, one mole of copper oxide makes one mole of copper sulfate.

Clearly the copper oxide is in excess, so the sulfuric acid is the limiting reactant, because you would need 0.125 moles of the acid to react with all 10 g of the copper oxide.

 

(b) Calculating the theoretical maximum yield of copper sulfate from the limiting reactant

This part of the calculation MUST be based on the limiting reactant (0.10 mol H2SO4)

From the equation, reading in molar quantities, one mole of sulfuric acid makes one mole of copper sulfate (0.1).

Formula mass copper sulfate = 64 + 32 + (4 x 16) = 160

mass = moles x formula mass = 0.10 x 160 = 60g CuSO4

 

Example 3. of a limiting reactant - making hydrogen from magnesium and hydrochloric acid

Suppose we take a strip of magnesium weighing 0.36g and dissolve it in 50 cm3 of hydrochloric acid of molarity 1.0 mol/dm3..

The equation is: Mg(s)  +  2HCl(aq)  ===>  MgCl2(aq)  +  H2(g)   and  Ar(Mg) = 24

 

(a) Which is the limiting reactant?

To calculate the limiting reactant we need to compare the number of moles of each reactant, AND then the relative reaction mole ratio to see which one is in excess - take care, you must not automatically just compare the actual number of moles of each reactant, you must consult the equation too!

mol Mg = mass Mg / Ar(Mg) = 0.36 / 24 = 0.015 mol Mg

mol HCl = [molarity HCl] x [volume of HCl(aq) in dm3]   (dm3 = cm3 / 1000)

mol HCl = 1.0 x 50 / 1000 = 0.05 mol HCl

According to the equation each mole of Mg requires two moles of HCl to complete the reaction.

Therefore 0.015 mol Mg requires 0.030 mol of HCl to react completely.

Since the moles of HCl required (0.03) is exceeded by the moles of HCl used (0.05),

the magnesium must be the limiting reactant

 

(b) What is the maximum volume of hydrogen gas you can make in this experiment from the limiting reactant?

This part of the calculation MUST be based on the limiting reactant (0.015 mol Mg)

Volume of 1 mol of gas = 24 dm3 at RTP (room temperature and pressure)

From the equation, reading in molar quantities, each mole of magnesium gives one mole of hydrogen gas,

so mol Mg used = mol H2 formed = 0.015 (limiting moles)

volume H2 = mol H2 x 24

= 0.015 x 24 = 0.36 dm3 (360 cm3)

 

Example 4. of a limiting reactant - preparing carbon dioxide from limestone and acid

2g of calcium carbonate granules (limestone) was dissolved in 20 cm3 of hydrochloric acid of concentration 0.50 mol/dm3.

The reaction is: CaCO3(s)  +  2HCl(aq)  ===>  CaCl2(aq)  + H2O(l)  +  CO2(g)

Atomic masses: Ca = 40; C = 12; O = 16; H = 1, Cl = 35.5; but you might not need all these Ar values!

 

(a) Which is the limiting reactant?

Formula mass CaCO3 = 40 + 12 + (3 x 16) = 100

mol CaCO3 = 2 / 100 = 0.02

mol HCl = [molarity HCl] x [volume of HCl(aq) in dm3]   (dm3 = cm3 / 1000)

mol HCl = 0.5 x 20 / 1000 = 0.01 mol

From the equation one mole of calcium carbonate requires two moles of hydrochloric acid to complete the reaction. Therefore the 0.02 mol of CaCO3 needs 0.04 mol of HCl to react completely.

Since only 0.01 mol of HCl was used in the mixture, clearly the calcium carbonate is in great excess and the hydrochloric acid is the limiting reactant.

Note: At the end of the reaction, indicated by no more 'fizzing', you would see white unreacted granules of limestone at the bottom of the flask.

 

(b) What is the maximum volume of carbon dioxide gas that can be obtained using this mixture from the limiting reactant?

This part of the calculation MUST be based on the limiting reactant (0.01 mol HCl)

From the equation, reading in molar quantities, two moles of hydrochloric acid produces one mole of carbon dioxide (so the moles of HCl will only make half the amount of moles of CO2).

Therefore mole CO2 = mol HCl / 2 = 0.01 / 2 = 0.005

volume CO2 = mol CO2 x 24 dm3

= 0.005 x 24 = 0.12 dm3  (120 cm3)

 


Above is typical periodic table used in GCSE science-chemistry specifications in doing chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

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OTHER CALCULATION PAGES

  1. What is relative atomic mass?, relative isotopic mass and 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)

  6. 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

  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-products)

  11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

  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

  15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

  16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

  17. Radioactivity & half-life calculations including dating materials

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