How many grams of NaAlO2 can be obtained from 4.46 g of AlCl3 according to the following balanced equation? AlCl3 + 4NaOH → NaAlO2 + 3NaCl + 2H2O

Answer 1

Molecular mass for AlCl₃ = 133.34 g/mol

Mol of AlCl₃ = (Mass of AlCl₃ / Molecular mass of AlCl₃) = (4.46 g) / (133.34 g/mol) = 0.0334 mol

Based on the balanced equation stoichiometric ratio of AlCl₃ : NaAlO₂ = 1:1

So,

Mol of NaAlO₂= 0.0334 mol

Molecular mass for NaAlO₂ = 81.97 g/mol

Mass of NaAlO₂ = (Mol of NaAlO₂ ) * (Molecular mass of NaAlO₂) = 0.0334 mol * 81.97 g/mol = 2.74 g