Answer:
- H₂SO₄: 6.99 g
- Pb(CH₃COO)₂ : 0
- PbSO₄: 9.31 g
- mol CH₃COOH: 3.57 g
Explanation:
1) Word equation
sulfuric acid + lead(II) acetate → lead(II) sulfate + acetic acid
2) Chemical equation (balanced)
H₂SO₄ (aq) + Pb(CH₃COO)₂ (aq) → PbSO₄ (s) + 2CH₃COOH (aq)
3) Mole ratios
1 mol H₂SO₄ : 1 mol Pb(CH₃COO)₂ : 1 mol PbSO₄ : 2 mol CH₃COOH
4) Molar masses:
- H₂SO₄: 98.079 g/mol
- Pb(CH₃COO)₂ : 325.2880 g/mol
- PbSO₄: 303.26 g/mol
- mol CH₃COOH: 58.0791 g/mol
4) Calculate number of moles of each reactant:
Formula: number of moles = mass / molar mass
- H₂SO₄: 10.0 g / 98.079 g/mol = 0.102 mol
- Pb(CH₃COO)₂ : 10.0 g/ 325.2880 g/mol = 0.0307 mol
5) Limiting reactant:
Since the thoretical mole ratio is 1 : 1, only 0.0307 moles of each reactant may react.
6) Mole chart
H₂SO₄ Pb(CH₃COO)₂ PbSO₄ CH₃COOH
Start 0.102 0.0307
React 0.0307 0.0307 - -
End 0.0713 0 0.0307 2×0.0307 = 0.0614
7) Convert the final mole numbers to grams, using the molar masses.
Formula: mass in grams = number of moles × molar mass
- H₂SO₄: 0.0713mol × 98.079 g/mol = 6.99 g
- Pb(CH₃COO)₂ : 0
- PbSO₄: 0.0307 mol × 303.26 g/mol = 9.31 g
- mol CH₃COOH: 0.0614 mol × 58.0791 g/mol = 3.57 g
8) Check the mass conservation:
i) Start: 10.0 g + 10.0 g = 20.0 g
ii) End: 6.99 g + 9.31 g + 3.57 g = 19.9
The 0.01 g difference is due to round decimals, so you conclude the results are good.
