. Ca(OH)2 + H2SO4 CaSO4 + H2O a) How many grams (g) of H2SO4 are needed to react with 40 grams of Ca(OH)2? b) How much CaSO4 (in grams) will be formed? c) How much H2O (in grams) will be formed?
The Correct Answer and Explanation is:
To solve this problem, we need to follow stoichiometric steps based on the balanced chemical equation:
Balanced equation:Ca(OH)2+H2SO4→CaSO4+2H2O\text{Ca(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2\text{H}_2\text{O}Ca(OH)2+H2SO4→CaSO4+2H2O
Step-by-step Solutions:
Step 1: Molar masses (g/mol)
- Ca(OH)₂ = 40.08 (Ca) + 2×(16.00 + 1.01) = 74.10 g/mol
- H₂SO₄ = 2×1.01 + 32.07 + 4×16.00 = 98.09 g/mol
- CaSO₄ = 40.08 + 32.07 + 4×16.00 = 136.15 g/mol
- H₂O = 2×1.01 + 16.00 = 18.02 g/mol
a) How many grams of H₂SO₄ are needed to react with 40 grams of Ca(OH)₂?
Step 1: Find moles of Ca(OH)₂:40.00 g74.10 g/mol≈0.54 mol\frac{40.00\ \text{g}}{74.10\ \text{g/mol}} \approx 0.54\ \text{mol}74.10 g/mol40.00 g≈0.54 mol
Step 2: Use the mole ratio (1:1) from the balanced equation:0.54 mol of Ca(OH)2 reactswith 0.54 mol of H2SO40.54\ \text{mol of Ca(OH)}_2\ reacts with\ 0.54\ \text{mol of H}_2\text{SO}_40.54 mol of Ca(OH)2 reactswith 0.54 mol of H2SO4
Step 3: Convert to grams:0.54 mol×98.09 g/mol≈52.97 g0.54\ \text{mol} \times 98.09\ \text{g/mol} \approx 52.97\ \text{g}0.54 mol×98.09 g/mol≈52.97 g
Answer: 52.97 grams of H₂SO₄
b) How much CaSO₄ (in grams) will be formed?
Mole ratio of Ca(OH)₂ to CaSO₄ is 1:1, so 0.54 mol of CaSO₄ is formed.0.54 mol×136.15 g/mol≈73.52 g0.54\ \text{mol} \times 136.15\ \text{g/mol} \approx 73.52\ \text{g}0.54 mol×136.15 g/mol≈73.52 g
Answer: 73.52 grams of CaSO₄
c) How much H₂O (in grams) will be formed?
From the balanced equation, 1 mol of Ca(OH)₂ makes 2 mol of H₂O0.54 mol Ca(OH)2×2=1.08 mol H2O0.54\ \text{mol Ca(OH)}_2 \times 2 = 1.08\ \text{mol H}_2\text{O}0.54 mol Ca(OH)2×2=1.08 mol H2O1.08 mol×18.02 g/mol≈19.46 g1.08\ \text{mol} \times 18.02\ \text{g/mol} \approx 19.46\ \text{g}1.08 mol×18.02 g/mol≈19.46 g
Answer: 19.46 grams of H₂O
Explanation:
This problem demonstrates stoichiometry, which is the calculation of reactants and products in chemical reactions. The equation tells us the mole relationships between substances. We use molar mass to convert grams to moles, apply mole ratios from the balanced equation, and convert moles of products or other reactants back to grams. Here, the 1:1 mole ratios between Ca(OH)₂, H₂SO₄, and CaSO₄ simplify the calculations. The only exception is H₂O, which has a 2:1 ratio with Ca(OH)₂, giving us twice as many moles of water per mole of base.
