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

94.0 g

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

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer