2. How many moles of Na2O (sodium oxide) are produced if 1.5 mol of Fe(iron) is
produced in the following reaction:
6Na + Fe2O3 - 3Na,O + 2Fe
3. If 4.3 moles of CaC2 (calcium carbonide) are used up in the reaction below, how
many grams of HO(water) are needed?
CaC2 + H20 - CaH2 +Ca(OH)?
Solution (2):
Balanced chemical equation:
6Na + Fe2O3 → 3Na2O + 2Fe
According to stoichiometry:
When 2 mol of Fe are produced, 3 mol of Na2O are produced.
When 1.5 mol of Fe are produced, X mol of Na2O are produced.
Therefore,
Moles of Na2O = X = (1.5 mol Fe) × (3 mol Na2O / 2 mol Fe) = 2.25 mol Na2O
Answer (2): 2.25 moles of Na2O (sodium oxide) are produced.
Solution (3):
Balanced chemical equation:
CaC2 + 2H2O → C2H2 + Ca(OH)2
According to stoichiometry:
1 mol of CaC2 reacts with 2 mol of H2O
Thus, 4.3 mol of CaC2 react with:
(4.3 mol CaC2) × (2 mol H2O / 1 mol CaC2) = 8.6 mol H2O
The molar mass of water (H2O) is 18.0 g/mol
Therefore,
Mass of H2O = (8.6 mol H2O) × (18.0 g H2O / 1 mol H2O) = 154.8 g H2O
Answer (3): 154.8 grams of H2O (water) are needed.
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