Answer to Question #198962 in Chemistry for Hsn

Question

Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

During an experiment 50g mass of Na2CO3 react with 0.5 mole of HCl. Calculate the actual mass of water produced.

Answer 1

The number of moles of Na₂CO₃ equals:

n(Na₂CO₃) = m(Na₂CO₃) / Mr(Na₂CO₃) = 50 g / 106 g/mol = 0.5 mol

Since the number of moles of both reactants are equal, 0.5 moles of water are produced during the reaction. From here, the mass of water equals:

m(H₂O) = n(H₂O) × Mr(H₂O) = 0.5 mol × 18 g/mol = 9 g

Answer: 9 g