Answer to Question #331773 in General Chemistry for matghjb

Question

6NaNO3 + Al2O3 → 2Al(NO3)3 + 3Na2O

If 82.1g NaNO3 reacts with 100g Al2O3, what is the limiting reactant?

Answer 1

6NaNO₃ + Al₂O₃ → 2Al(NO₃)₃ + 3Na₂O

n(NaNO₃) = m(NaNO₃)/Mr(NaNO₃) = 82.1 g / 84.995 g/mol = 0.966 mol

n(Al₂O₃) = m(Al₂O₃)/Mr(Al₂O₃) = 100 g / 101.96 g/mol = 0.981 mol

6 mol NaNO₃ react with 1 mol Al₂O₃

0.966 mol react with x mol Al₂O₃

x = 0.966 / 6 = 0.161 mol

NaNO₃ is limiting reactant