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.

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

explanation: no but i do not usually comment