2HCl + Na2CO3 --> 2NaCl + CO2 + H2O

What volume of 0.2moldm-3 solution of sodium carbonate is required to react completely with 50cm3 0.5moldm-3 hydrochloric acid?A.31.5cm3

B.62.5cm3

C.125cm3

D.250cm3

Explanation:

No. of moles of HCl = (50x0.5)/1000

= 0.025

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

= 62.5

Therefore, volume of sodium carbonate required is 62.5 cm3.

Question asked by Ryane.

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