Question

Aqueous hydrochloric acid HCl reacts with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O . If 0.102g of water is produced from the reaction of 0.73g of hydrochloric acid and 0.58g of sodium hydroxide, calculate the percent yield of water.

Answer 1

HCl + NaOH ---> NaCl + H2O

Molar mass of HCl = 36.46 g/mol

0.73 g / 36.46 g/mol => 0.02 mol of HCl

Molar mass of NaOH = 40 g/mol

0.58 g / 40 g/mol => 0.0145 mol of NaOH

for complete neutralization of 0.02 mol of HCl, It would require 0.02 mol of NaOH, But only 0.0145 mol of NaOH is available.

Hence NaOH is limitting reagent.

Therotical yield of H2O = 0.0145 mol

Molar mass of H2O = 18 g/mol

0.0145 mol * 18 g/mol => 0.261 g of water

% Yield of water = (0.102 g / 0.261 g )*100 => 39 %