Answer:
148 g H₂SO₄
General Formulas and Concepts:
Chemistry - Stoichiometry
- Reading a Periodic Table
- Balancing RxN's
- Using Dimensional Analysis
Explanation:
Step 1: Define
RxN: Ca(OH)₂ + H₂SO₄ → CaSO₄ + H₂O
Given: 3.01 moles H₂O
Step 2: Balance RxN
Ca(OH)₂ + H₂SO₄ → CaSO₄ + 2H₂O
- Need the same amount of O's and H's on both sides
Step 3: Define conversions
Molar Mass of H - 1.01 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂SO₄ - 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol
Step 4: Stoichiometry
[tex]3.01 \ mol \ H_2O(\frac{1 \ mol \ H_2SO_4}{2 \ mol \ H_2O} )(\frac{98.09 \ g \ H_2SO_4}{1 \ mol \ H_2SO_4} )[/tex] = 147.625 g H₂SO₄
Step 5: Check
We are given 3 sig figs. Follow sig fig rules.
147.625 g H₂SO₄ ≈ 148 g H₂SO₄
