Express of `CO_(2)` is passed through 50 mL of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonated was completely neutralized with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (At mass of carbon = 40)

Options

A. 200 `cm^(3)`
B. `500 cm^(3)`
C. `400 cm^(3)`
D. `300 cm^(3)`

Answer

B

Explanation

`Ca(OH)_(2) + CO_(2) to CaCO_(3) + H_(2)O`
Moles of `Ca(OH)_(2)` present in 50 mL of 0.5 M solution
`=0.5/1000 xx 50 =2.5 xx 10^(-2)` mol
Moles of `CaCO_(3)` formed `=2.5 xx 10^(-2)` mol
`CaCO_(3) + 2HCl to CaCl_(2) + H_(2)O + CO_(2)`
Moles of HCl required
`=2 xx 2.5 xx 10^(-2)` mol
`5.0 xx 10^(-2)` mol
Volume of 0.1 N HCl required
`=0.1/1000 xx x = 5.0 xx 10^(-2)`
`x = (5.0 xx 10^(-2) xx 1000)/0.1 = 500 mL`