Answer to Question #82352 in General Chemistry for Payton Pechulis

Question

Calculate the number of grams of lead(II) acetate present in the mixture after the reaction is complete.

Answer 1

Solution:

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 5.90 g of sulfuric acid and 5.90 g of lead(II) acetate are mixed.

Calculate the number of grams of lead(II) acetate present in the mixture after the reaction is complete

H2SO4 + Pb(CH3COO)2 → PbSO4↓ + 2CH3COOH

M(H2SO4) = 2 + 32 + 64 = 98 g/mol;

M(Pb(CH3COO)2) = 207 + (12 + 3 + 12 + 32)2 = 325 g/mol

n(H2SO4) = m/M = 5.9 g / 98 g/mol = 0.060 mol

n(Pb(CH3COO)2) = m/M = 5.9 g / 325 g/mol = 0.018 mol

n(Pb(CH3COO)2)  n(H2SO4)

m(Pb(CH3COO)2) = 0 g after the reaction is complete