The mole and chemical equations
Francine Taylor-Campbell, Contributor

A section of the audience, which attended the opening of Science Awareness Week at Bridgeport High School on Monday, February 20 under the theme: 'Portmore, keeping it green'. - Anthony Minott Photo

Please note:

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 0 = 16 H = 1 N = 14)

PbCO3(s) + 2HNO3 (aq) = Pb (NO3)2(aq) + H2O (1) + CO2 (g)

1 mole 2 moles 1 mole 1 mole 1 mole

267g 2*63g 331g 18g 44g = 24 dm³ at RTP.

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 HNO 3 = 0.5 mol Pb
(NO3)2

2 moles HNO3 = 24 dm³ CO2 at RTP

1 mole HNO3 = 12 dm³ CO2
at RTP.

Wolmer's' Adrian Blake shoots over St. Catherine's Noel Wright in the KFC ISSA High School Basketball Southern Conference action at the Mico College yesterday. St. Catherine won the game 59-47, to take a 1-0 series lead in the best-of-three finals. - Ricardo Makyn Photo

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 and 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 dm of 2 mol/dm³ HNO3

i.e. 267g PbCO3 react with 1000 cm³ HNO3

26.7 will react with 100 cm³ HNO3

Excess PbCO(3) = 30 - 26.7 = 3.3g

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

Ans: 267g PbCO3 = 24 dm³ CO(2) at RTP

26.7g PbCO(3) = 2.4 dm(3) 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 Fes + 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 H2SO(4) would react to produce 4.8 dm3 H2 at RTP?

* Francine Taylor-Campbell is an independent contributor.