The mole and chemical equations
Francine Taylor-Campbell, Contributor

SYLLABUS REQUIREMENT

• Apply the mole concept to equations.
• Write balanced equations including state symbols to represent chemical reactions.
• Perform calculations involving the mole.

BASIC FACTS

• To perform calculations based on chemical reactions, an equation must first be written, then balanced, so that the mole concept can be applied.

CALCULATIONS BASED ON EQUATIONS

Eg The reaction ( Pb = 207 C = 12 O = 16 H = 1 N = 14 )

PbCO_3(s) + 2HNO_{3 (aq)} = Pb(NO₃)_{2 (aq)} + H₂O _{(l )} + CO_{2 (g)}

Questions

1. How many moles of nitric acid are needed to obtain 0.5 moles of lead nitrate, what volume

of carbon dioxide is obtained in the same experiment (at RTP)?

Ans

2 moles HNO₃ = 1 mole Pb(NO₃)₂
1 mole HNO₃ = 0.5 mol Pb(NO₃)₃

2 moles HNO₃ = 24 dm³ CO₂ at RTP
1 mole HNO₃ = 12 dm³ CO₂ at RTP

2. If the nitric acid contains 2 moles in one dm³

(2 mol/dm³), what volume of nitric acid (in cm³) would be needed in Q1?

Ans

2 moles HNO₃ are contained in 1,000 cm³ solution
0.5 moles HNO₃ are contained in 250 cm³ solution

3. How many grams of lead nitrate could be obtained from 53.4g of lead carbonate reacting with an excess of acid.

Ans

267g PbCO₃ = 331g Pb(NO₃)₂

53.4 PbCO₃ = (331*53.4)/267 = 66.2g Pb(NO₃)₂

Note: In Q3, the limiting reagent is lead carbonate; all of it reacts. In preparing lead nitrate in

the laboratory, an excess of lead carbonate is used, hence the limiting reagent is nitric acid.

4. 30g PbCO₃ were reacted with 100 cm3 of 2 mol/dm3 HNO3. When the reaction was complete,

what mass of PbCO₃ remained unreacted?

Ans

From the equation, 267g PbCO₃ react with 2 moles HNO₃

267g PbCO₃ react with 1 dm³ of 2mol/dm³ HNO₃

ie, 267g PbCO₃ react with 1000 cm³ HNO₃

26.7g will react with 100 cm³ HNO₃

Excess PbCO₃ = 30 - 26.7 = 3.3g

5. What volume of CO2 (at RTP) is produced in the experiment in Q4?

Ans

267g PbCO₃ = 24 dm³ CO₂ at RTP
26.7g PbCO₃ = 2.4 dm³ CO₂ at RTP.

Now attempt this question.

6. Iron sulphate was prepared by reacting an excess of iron with 100 cm³ of 1 mol/dm³

sulphuric acid. (Fe = 56 S = 32 H = 1 O = 16)

Equation Fe_(s) + H₂SO₄ (aq) = FeSO_{4 (aq )}+ H₂(g)

a. What mass of iron reacted?

b. What mass of FeSO4 could be produced?

c. What volume of hydrogen at RTP would be obtained?

d. When crystalline FeSO4.7H2O is obtained, what mass of this could be obtained?

e. What volume of 1 mol/dm3 H2SO4 would react to produce 4.8 dm3 H2 at RTP?

ANSWERS to Question 6

a. 5.6g of iron.

b. 15.2g FeSO₄.

c. 2.4 dm3 of hydrogen.

d. 27.8g FeSO₄.7H₂O crystals.

e. 200 cm3 of H₂SO₄.

Francine Taylor-Campbell is an independent contributor. Send questions and comments to kerry-ann.hepburn@gleanerjm.com