Fundamental Chemistry: Theory and Practice Carnegie College
Topic 5 – Calculations from Balanced Chemical Equations
Acknowledgements
No extract from any source held under copyright by any individual or organisation has been included in this Booklet.
The author would like to thank the following people for their assistance in the development of these materials:
Alan McDowall
Contents
Topic 5 – Calculations from Balanced Chemical Equations 1
Calculations Using Balanced Chemical Equations 1
Example 1 2
Example 2 4
Calculations Involving Excess Reagent 8
The Calculations from Balanced Chemical Equations Progress Checklist 12
Answers to SAQs 13
© Carnegie College DH2K 34
Fundamental Chemistry: Theory and Practice Carnegie College
Topic 5 – Calculations from Balanced Chemical Equations
Topic 5 – Calculations from Balanced Chemical Equations
By the end of this section you should be able to:
· Carry out calculations using balanced chemical equations
· Carry out calculations involving excess reagent
Calculations Using Balanced Chemical Equations
In the last booklet we introduced some basic chemical principles. In this section we are going to learn how using our basic knowledge useful calculations can be carried out.
We have learned that chemical reactions can be shown in a shortened form as chemical equations and that chemical equations tell us the amounts of substances that are taking part in the reaction.
Consider the reaction below
NaOH(aq) + HCl(aq) NaCl(aq) + H2O
This reaction represents the neutralisation of an acid by an alkali. It tells us that when 1 mole of sodium hydroxide reacts with 1 mole of hydrochloric acid 1 mole of sodium chloride and 1 mole of water are produced.
If we work out the gram formula mass of each of the compounds we are able to see the masses that react and the masses of product that will be formed.
NaOH + HCl NaCl + H2O
gfm 40g 36∙5g 58∙5g 18g
Our ability to do this allows us to calculate the mass of product that can be formed from given masses of reactants, or the mass of reactant required to produce a given mass of product. In such calculations it is important both to identify and to concentrate on only the chemicals we are asked to calculate masses for and the chemicals for which we have been given masses.
Example 1
Calculate the mass of carbon dioxide produced when 10g of calcium carbonate is decomposed by heating.
The best way to tackle problems like these is in a series of steps.
Step 1: Write a balanced chemical equation
CaCO3 CaO + CO2
Step 2: Identify the chemicals mentioned in the question
CaCO3 CaO + CO2
We can ignore CaO as this is not mentioned in the question
Step 3: Highlight the molar ratio of the compounds
CaCO3 CaO + CO2
1 mole 1 mole
Step 4: Convert the number of moles to gfm.
CaCO3 CaO + CO2
100g 44g
10g ?
Step 5: Complete the calculation
This is an easy calculation and you should be able to see that the answer is 4.4g. But how do we do the calculation if the answer is not so straightforward?
There are 2 ways this can be done it is important that you find the way that suits you best and stick to it!
Method 1
Use cross multiplication
Method 2
Use the knowledge triangles
Calculate how many moles of CaCO3 are contained in 10g.
Now look at the molar ratio from the balanced chemical equation and we see that 1mole CaCO3 decomposes to give 1 mole of CO2; so 0∙1 mole will decompose to give 0∙1moles of CO2. Use the knowledge triangles again to complete your calculation.
Example 2
Calculate the mass of magnesium oxide formed if 12g of magnesium is burned in excess oxygen. This illustrates another point that we need to be aware of; when a question says “in excess” this means that there is more than enough for the reaction to go to completion and indicates that we can ignore the chemical in excess in our calculations. In the example above this means we can ignore oxygen.
Follow the steps above to complete the calculation
Step 1: Write a balanced chemical equation
2Mg + O2 2MgO
Step 2: Identify the chemicals mentioned in the question
Remember oxygen is in excess and can be ignored
2Mg + O2 2MgO
Step 3: Highlight the molar ratio of the compounds
2Mg + O2 2MgO
2 moles 2 moles
Step 4: Convert the number of moles to gfm.
2Mg + O2 2MgO
48∙6g 80∙6g
12g ?
Step 5: Complete the calculation
Method 1
Method 2
Now try SAQ 1 on the following page
1
Now try the following examples.
1 Calculate the weight of hydrochloric acid that would be required to react with 80 g of sodium hydroxide?
HCl + NaOH NaCl + H2O
2 Calculate the mass of magnesium that would react with 9.125 g of hydrochloric acid?
Mg + 2HCl MgCl2 + H2
3 Calculate the mass of water produced when 80 g of methane (CH4) is completely burned in air (completely burned in air is another way of telling us oxygen is present in excess).
CH4 + 202 CO2 + 2H2O
4 Calculate the weight of oxygen required to completely burn 120 g of carbon.
C + O2 CO2
SAQ 1 continues over the page
SAQ 1 continued
5 Determine the mass of sodium sulphate formed when 10.0 g of sodium hydroxide is completely neutralised by sulphuric acid.
2NaOH + H2SO4 Na2SO4 + 2H2O
6 Calculate the mass of sulphur dioxide produced when 500 g of sulphur is completely oxidised.
S + O2 SO2
7 Calculate the mass of calcium carbonate formed when 500 g of calcium chloride reacts with excess sodium carbonate.
CaCl2 + Na2CO3 CaCO3 + 2NaCl
Check your answers with those given at the end of this booklet.
Calculations Involving Excess Reagent
It is also possible to calculate which reactant is in excess and so work out the actual mass of product that can theoretically be formed.
Consider the example below.
During a reaction 1.31g of zinc reacted with 30cm3 of 2moll-1 hydrochloric acid.
(a) Show by calculation which reactant is in excess
(b) Calculate the mass of hydrogen gas produced
Once again balanced chemical equations are essential in this type of work.
Step 1: Write a balanced chemical equation
Zn + 2HCl ZnCl2 + H2
Step 2: We are told both the quantities of zinc and of hydrochloric acid so these must both be included in the calculation. We are asked to calculate the quantity of hydrogen so this will also be in the calculation. The only thing we can remove is the ZnCl2 as this is not important in this example.
For step 2 we need to look at the molar ratios.
Zn + 2HCl ZnCl2 + H2
1 mole 2moles 1 mole
Step 3: Calculate the values for the known moles
For zinc n = m ÷ gfm n = 1∙31 ÷ 65.4 = 0∙02moles
For HCl n = C x V = 2 x 0∙03 = 0∙06 moles
From the balanced equation we can see that 1 mole of zinc will react with 2 moles of hydrochloric acid. So 0∙02 moles of zinc will react with 0∙04 moles of HCl. Since we have more than 0∙04 moles of acid hydrochloric acid must be in excess.
Step 5: Calculate the mass of H2 produced.
We do this by ignoring the reagent in excess in this case the HCl and looking again at the balanced equation.
Zn + 2HCl ZnCl2 + H2
1mole 1 mole
0∙02moles 0∙02moles
This allows us to finish off the calculation by calculating the mass of hydrogen produced
Now try SAQ 2 on the following page
2
Now try the following examples.
For each example show by calculation which reactant is present in excess before completing the question asked?
1 6∙22g of lead is added to 50cm3 of 1moll-1 hydrochloric acid. Calculate the mass of hydrogen gas evolved.
Pb(s) + 2HCl(aq) PbCl2(aq) + H2(g)
2 Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)
Calculate the theoretical mass of lead iodide precipitate formed given that 120cm3 of a 0∙2moll-1 lead(II)nitrate solution was added to 200cm3 of 0.25moll-1 potassium iodide solution.
3 Using the equation below calculate the mass of aluminium hydroxide that would be precipitated in this reaction given that 40cm3 of 0∙5moll-1 aluminium nitrate was added to 200cm3 of a 0∙4moll-1 solution of sodium hydroxide.
Al(NO3)3(aq) + 3NaOH(aq) Al(OH)3(s) + 3NaNO3(aq)
SAQ 2 continues on the following page
SAQ continued
4 H2SO4(aq) + BaCl2(aq) BaSO4(s) + 2HCl(aq)
Calculate the mass of barium sulphate precipitated when 20cm3 of 0∙2moll-1 sulphuric acid is added to 50cm3 of 0∙1moll-1 barium chloride solution.
Check your answers with those given at the end of this booklet.
The Calculations from Balanced Chemical Equations Progress Checklist
Tick the boxes only if you can:
Topic / Understand?Carry out calculations from balanced chemical equations
Carry out equations involving excess reagent
Answers to SAQs
Answer to SAQ 1
1) 73g
2) 3∙04g
3) 180g
4) 320g
5) 17∙76g
6) 998∙44g
7) 450∙45g
Answer to SAQ 2
1) Pb in excess 0∙05g
2) KI in excess 11∙064g
3) NaOH in excess 1∙56g
4) BaCl2 0∙93g
13
© Carnegie College DH2K 34