Leaddioxide + Conc. Hydrochlaric acid = Lead
chloride + Water + Chloride
Answers
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