`CaCO_(3)` reacts with HCl to produce `CaCl _(2), CO_(2)and H _(2)O.` The approximate mass (in g) of `CaCO_(3)` required t react completely with 25 mL of `0.75` M HCl is (atomic mass of `Ca = 40, C =12 O =16, Cl =35.5and H =1)`

Options

A. 94
B. `9.4`
C. `0.94`
D. `0.094`

Answer

C

Explanation

`2HCl+ CaCO_(3) to CaCl_(2) + CO _(2) + H_(2)O`
25 mL of `0.75 N, HCl =0.01875` mle of HCl . As per reaction twice the amount of HCl is needed in comparsion to `CaCO_(3).`
Thus, amount of `CaCO_(3)` neutralised.
`= (0.01875)/(2) = 0.009375` mol
`0.009375 mol xx 100 g //mol = 0.9383 g of CaCO_(3) = 0.94g`