Question

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

Answer 1

Hence, correct answer is option A.

Step-by-step explanation:

Mass of calcium carbonate = 3.25 g

Moles of calcium carbonate =
`(3.25 g)/(100 g/mol)=0.0325 mol`

`2 HNO_3(aq) + CaCO_3(s)\rightarrow H_2O(l) + CO_2(g) + Ca(NO_3)_2(aq)`

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 :

`(2)/(1)* 0.0325 mol=0.0650 mol`

Moles of nitric acid = 0.0650 mol

Volume of the nitric acid solution = V

Molarity of the nitric solution = 0.300 M

`Molarity=(Moles)/(Volume (L))`

`0.300 M=(0.0650 M)/(V)`

`V=(0.0650 mol)/(0.300 M)=0.216 L`

1 L = 1000 mL

0.216 L = 0.216 × 1000 mL = 216 mL

Hence, correct answer is option A.