How many milliliters of 0.300 M nitric acid are required to react completely with 3.25 g of calcium carbonate?
2 HNO3(aq) + CaCO3(s) H2O(l) + CO2(g) + Ca(NO3)2(aq)

A. 216
B. 32.5
C. 0.216
D. 108
E. 1950

Hence, correct answer is option A.

Explanation:

Mass of calcium carbonate = 3.25 g

Moles of calcium carbonate =

According to reaction, 1 mole of calcium carbonate reacts with 2 moles of nitric acid then 0.0325 moles of calcium carbonate will react with :

Moles of nitric acid = 0.0650 mol

Volume of the nitric acid solution = V

Molarity of the nitric solution = 0.300 M

1 L = 1000 mL

0.216 L = 0.216 × 1000 mL = 216 mL

Hence, correct answer is option A.

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