Solution

Step 1 Calculate the mole of HCl in excess
Mole = volume X concentration = 0.027 X 0.1 = 0.0027

Step 2 Calculate the mole of HCl that reacted with the calcium carbonate.

Mole of HCl originally added - mole of HCl in excess = mole of HCl reacted
0.05 X0.1 - 0.0027 = 0.005 - 0.0027 =0.0023 mole

Step 3 Using the equation below calculate the number of mole of calcium carbonate required to react completely with 0.0023 mole of HCl

0.0023 /2 = 0.00115 mole of calcium carbonate is present.

Step 4 Find the percentage by mass of the calcium carbonate.

Mass of calcium carbonate = 0.00115 X 100 = 0.12grams
(0.115 / 1.532 ) X 100 = 7.51%

 

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