Answer to Question #200131 in Chemistry for marianna

Question

If you start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed? 

Answer 1

Pb(NO₃)₂ + 2NaI = 2NaNO₃ + PbI₂

n = m / M

M (Pb(NO₃)₂) = 331.2 g/mol

M (NaI) = 150 g/mol

M (NaNO₃) = 85 g/mol

n (Pb(NO₃)₂) = 25 / 331.2 = 0.08 mol

n (NaI) = 15 / 150 = 0.1 mol

According to the equation, n (NaI) = 2 x n (Pb(NO₃)₂). From the given reactant quantities, NaI is the limiting reactant. n (NaNO₃) = n (NaI). The amount of NaNO₃ to be formed is:

m (NaNO₃) = n x M = 0.1 x 85 = 8.5 g