Answer to Question #305506 in General Chemistry for Mcquay

Question

Pb(NO₃)₂ + 2 NaI à PbI₂ + 2 NaNO₃ 

If 100 grams of lead (II) nitrate were used in the reaction, how many grams of sodium iodide were used for a complete reaction? 

Answer 1

Pb(NO₃)₂ + 2 NaI = PbI₂ + 2 NaNO₃

According to the equation, n (NaI) = 2 x n (Pb(NO₃)₂).

n = m / M

M (Pb(NO₃)₂) = 331.2 g/mol

M (NaI) = 149.9 g/mol

n (Pb(NO₃)₂) = m / M = 100 / 331.2 = 0.3 mol

n (NaI) = 2 x n (Pb(NO₃)₂) = 2 x 0.3 = 0.6 mol

m (NaI) = n x M = 0.6 x 149.9 = 90 g