How Many grams of NaN3 can be made by the reaction of 3.50g of NaNH2 and 3.50 g of NaNO3?

Molar mass of NaNH₂ = (22.99 + 14.01 + 1.008×2) g/mol = 39.02 g/mol
Molar mass of NaNO₃ = (22.99 + 14.01 + 16.00×3) g/mol = 85.00 g/mol
Molar mass of NaN₃ = (22.99 + 14.01×3) g/mol = 65.02 g/mol

Initial no. of moles of NaNH₂ = (3.50 g) / (39.02 g/mol) = 0.0897 mol
Initial no. of moles of NaNO₃ = (3.50 g) / (85.00 g/mol) = 0.0412 mol

3NaNH₂ + NaNO₃ → NaN₃ + 3NaOH + NH₃
Mole ratio NaNH₂ : NaNO₃ = 3 : 1

If 0.0897 mol of NaNH₂ completely reacts :
No. of moles of NaNO₃ needed = (0.0897 mol) × (1/3) = 0.0299 mol < 0.0412 mol
Hence, NaNO₃ is in excess, and NaNH₂ is the limiting reactant (completely reacts)

According to the above equation, mole ratio NaNH₂ : NaN₃ = 3 : 1
No. of moles of NaNH₂ reacted = 0.0897 mol
No. of moles of NaN₃ made = (0.0897 mol) × (1/3) = 0.0299 mol
Mass of NaN₃ made = (65.02 g/mol) × (0.0299 mol) = 1.94 g