Answer to Question #305489 in General Chemistry for Mcquay

Question

Pb(NO₃)₂ + 2 NaI à PbI₂ + 2 NaNO₃  

7)   How many grams of the product were formed in no. 6 if the percent yield is 80%?

Answer 1

Pb(NO₃)₂ + 2 NaI = PbI₂ + 2 NaNO₃ 

If we had 5 g of Pb(NO₃)₂

5 g Pb(NO₃)₂ - x g PbI₂

331.2 g/mol Pb(NO₃)₂ - 461.01g/mol PbI₂

x = 461.01 × 5 / 331.2 = 6.96 g

m(PbI₂) = 0.8 × 6.96 g = 5.57 g

5 g Pb(NO₃)₂ - y g NaNO₃

331.2 g/mol Pb(NO₃)₂ - 2 mol × 84.99 g/mol NaNO₃

y = 2 × 84.99 × 5 / 331.2 = 2.57 g

m(NaNO₃) = 0.8 × 2.57 g = 2.06 g