Answer:a) First, we write the chemical reaction and we CALCULATE the relative molecular mass (or

atomic mass if it's an element) of each SUBSTANCE in the reaction:

PbO2 + 4 HCl → PbCl2 + Cl2 + 2 H2O

PbO2 Mm = 1 · 207,2 + 2 · 16,0 = 239,2 u (

1

)

HCl Mm = 1 · 1,0 + 1 · 35,5 = 36,5 u

PbCl2 Mm= 1 · 207,2 + 2 · 35,5 = 278,2 u

Cl2 Mm= 2 · 35,5 = 71,0 u

H2O Mm= 1 · 1,0 + 2 · 16,0 = 18,0 u

Mm 239,2 36,5 278,2 71,0 18,0

PbO2 + 4 HCl → PbCl2 Cl2 2 H2O

b) Second, we calculate the amount of substance (n) with the given data in the statement.

n = 31,2 g PbO2 · 1 mol PbO2 / 239,2 g PbO2 = 0,130 moles PbO2

c) Third, we balance the chemical equation.

Chemical reaction: PbO2 + 4 HCl → PbCl2 + Cl2 + 2 H2O

Pb 1 1

Cl 2 2 4

H 4 1

O 2 2

The equation is already balanced.

PbO2 + 4 HCl → PbCl2 + Cl2 + 2 H2O

d) Then, we collect all the information:

Mm 239,2 36,5 278,2 71,0 18,0

Stoichiometry 1 mol 4 moles 1 mol 1 mol 2 moles

PbO2 + 4 HCl → PbCl2 Cl2 2 H2O

Amount of

substance

0,130

moles

x moles y moles z moles t moles

1 unified atomic mass unit, symbol (u)

e) Fourth, we calculate the amount of the substances (n) in the chemical reaction:

x moles HCl = 0,130 moles PbO2 · 4 mol HCl / 1 mol PbO2 = 0,520 moles HCl

y moles PbCl2 = 0,130 moles PbO2 · 1 mol PbCl2 / 1 mol PbO2 = 0,130 moles PbCl2

z moles Cl2 = 0,130 moles PbO2 · 1 mol Cl2 / 1 mol PbO2 = 0,130 moles Cl2

t moles H2O = 0,130 moles PbO2 · 2 moles H2O / 1 mol PbO2 = 0,260 moles H2O

f) We collect the information again:

Mm 239,2 36,5 278,2 71,0 18,0

Stoichiometry 1 mol 4 moles 1 mol 1 mol 2 moles

PbO2 + 4 HCl → PbCl2 Cl2 2 H2O

Amount of

substance

0,130

moles

0,520

moles

0,130

moles

0,130

moles

0,260

moles

g) Then, we can answer the questions in the statement by USING the amounts of substances:

What mass of lead dichloride would be obtained from 37,2 g of PbO2?

(step e) There will be produced 0,130 moles PbCl2

The mass is: 0,130 moles PbCl2 · 278,2 g PbCl2 /1 mol PbCl2 = 36,17 g PbCl2

What mass of CHLORINE gas would be produced?

(step e) There will be produced 0,130 moles Cl2

The mass is: 0,130 moles Cl2 · 71,0 g Cl2 /1 mol Cl2 = 9,23 g Cl2