The mole and chemical equations
Francine Taylor-Campbell, Contributor

Dr. Hamadoun I. Toure, Secretary-General of the International Telecommunications Union, speaks with students, on Monday, April 16, while on a tour of the e-Learning project at Camperdown High School, Kingston. - Rudolph Brown/Chief Photographer

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

Calculations based on equations

E.g. the reaction (Pb = 207, C = 12, O = 16, H = 1, N = 14)

PbCO3(s) + 2HNO3(aq) = Pb(NO 3)2 (aq) + H2O(l) + CO2(g)

Exercises:

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)2

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 1000 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

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

That is, 267g PbCO3 react with 1000 cm³ HNO3

26.7g will react with 100 cm³ HNO3

Excess PbCO3 = 30 - 26.7 = 3.3g

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

Ans: 267g bCO3 = 24 dm³ CO2 at RTP

26.7g PbCO3 = 2.4 dm3 CO³ at RTP.

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) = FeSO4(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

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.