Answer to Question #305507 in General Chemistry for Mcquay

Question

Pb(NO₃)₂ + 2 NaI = PbI₂ + 2 NaNO₃ 

Suppose in the reaction in which 100 grams of lead(II) nitrate were all used up and only 100 grams of lead(II)iodide were obtained in the experiment, what is the percent yield? 

Answer 1

Pb(NO₃)₂ + 2 NaI = PbI₂ + 2 NaNO₃

According to the equation, n (PbI₂) = n (Pb(NO₃)₂).

n = m / M

M (Pb(NO₃)₂) = 331.2 g/mol

M (PbI₂) = 461 g/mol

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

n (PbI₂) = n (Pb(NO₃)₂) = 0.3 mol

m (PbI₂)_theoretical = n x M = 0.3 x 461 = 138.3 g

% (PbI₂) = (100 x 100) / 138.3 = 72.3%