Solutions of sulfuric acid and lead (ii) acetate react to form solid lead (ii) sulfate and a solution of acetic acid. if 10.0 g of sulfuric acid and 10.0 g of lead (ii) acetate are mixed and assuming the reaction goes to 100% completion, report the mass amount (in grams) of sulfuric acid, lead (ii) acetate, lead (ii) sulfate, and acetic acid. report your answers to 2 s. f. (hint: in other words, how much of each product is produced once the reaction is completed? how much of each reactant remains when the reaction is completed

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.