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.
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