Question

If 0.582 moles of zinc reacts with excess lead(IV) sulfate how many grams are zinc sulfate would be produced?

Answer 1

Answer: 94.0 g

Step-by-step explanation:

1) Reactants:

i) zinc (limiting)

ii) lead(IV) sulfate (excess); Pb(SO₄)₂

2) Chemical equation:

2 Zn + Pb(SO₄)₂ → 2 ZnSO4 + Pb

3) mole ratios

2 mol Zn: 1 mol Pb(SO₄)₂ : 2 mole ZnSO₄ : 1 mol Pb

4) Proportion

x / 0.582 mol Zn = 2 mole ZnSO₄ / 2 mol Zn

⇒ x = 1 × 0.582 mole ZnSO₄ = 0.582 mol ZnSO₄

5) Convert 0.582 mole ZnSO₄ to grams

mass in grams = number of moles × molar mass

molar mass ZnSO₄ = [65.39 + 32.065 + 4 × 15.999] g/mol = 161.451 g/mol

mass in grams = 0.582 mol × 161.451 g/mol = 93.96 g

Round to three significant figures: 94.0 g

Answer 2

From the balanced equation 2 moles of Zn make 2 moles of ZnSO4

You have 0.582 moles of zinc so you make 0.582 moles of ZnSO4.

Calculate the molar mass of ZnSO4. Multiply that by the moles of ZnSO4.

molar mass of ZnSO4 = 161.47 g/mol
zinc sulfate produced = 161.47 g/mol x 0.582 moles
= 93.97 grams