2HCl + Na2CO3 --> 2NaCl + CO2 + H2OWhat volume of 0.2moldm-3 solution of sodium carbonate is required to react completely with 50cm3 0.5moldm-3 hydrochloric acid?
No. of moles of HCl = (50x0.5)/1000
From the balanced equation given,
2 mol of HCl react completely with 1 mol of Na2CO3
Therefore, 0.025 mol of HCl will react completely with 0.025/2=0.0125 mol of Na2CO3
0.0125 = (0.2V)/1000
0.2V = 12.5
V = 12.5/0.2
Therefore, volume of sodium carbonate required is 62.5 cm3.
Question asked by Ryane.