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)rarr CaCO_(3)+H_(2)O`
50 mL of 0.5 `Ca(OH)_(2)` contain `Ca(OH)_(2)`
`=(0.5)/(1000)xx"50 mol"=0.025"mol"`
1 mol of `Ca(OH)_(2)` produces `CaCO_(3)` = 1 mol
`therefore 0.25` mole of `Ca(OH)_(2)` will produce `CaO_(3)`
= 0.025 mol
`CaCO_(3)+2HClrarr CaCl_(2)+H_(2)O+CO_(2)`
1 mole of `CaCO_(3)` is neutralised by HCl = 2 mol
`therefore 0.025` mol of `CaCO_(3)` will be neutralised by HCl
`=2xx0.025" mol" = 0.05 " mol"`
`" 1000 mL of 0.1 N HCl contain HCl = 0.1 g eq"`
`" = 0.1 mol"`
`"or 0.1 of HCl is present in HCl sol = 1000 mL"`
`therefore" 0.05 mol of HCl will be present in HCl sol."`
`=(1000)/(0.1)xx0.05mL="500 mL"`