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 mL`
B. `500 mL`
C. `400 mL`
D. `300 mL`

Answer

B

Explanation

`Ca(OH)_(2)+CO_(2)toCaCO_(3)+H_(2)O`
50 mL of 0.5 mL `Ca(OH_(2)) = (0.5)/(1000)xx50`mol = 0.025 mol
1 mol of `Ca(OH)_(2)` produces `CaCO_(3)=1` mol
`:.` 0.025 mol of `Ca(OH)_(2)` will produce `CaCO_(3) = 0.025` mole
`CaCO_(3)+2HCl toCaCl_(2)+H_(2)O+CO_(2)`
1 mole of `CaCO_(3)` is neutralized by HCl = 2 mol
=0.25 mol `CaCO_(3)` will be neutralized by HCl `=2xx0.025` mol = 0.05 mol
100 mL of 0.1 N HCl contain HCl = 1 eq. =0.1 mol
or 0.1 mol of HCl is present in HCl sol = 1000 mL
`:.` 0.05 mole of HCl is present in HCl sol
`:. (1000)/(0.1)xx0.05 mL = 500 mL`