What mass of `CaCO_(3)` is required to react completely with 25 ml of `0.75 MHCI`?

Options

A. `0.94 g`
B. `9.4 g`
C. `0.094 g`
D. `0.49 g`

Answer

A

Explanation

`CaCO_(3) + 2HCl to CaCl_(2) + CO_(2) + H_(2)`
25 mL of 0.75 M HCl
`= (25)/(1000)Lxx(0.75 mol L^(-1))`
`= 0.01875 mol`
Moles of `CaCO_(3)` required `= ("Moles of HCl")/(2)`
`= (0.01875)/(2) = 9.375xx10^(-3) mol`
Mass of `CaCO_(3)` required
`= 9.375xx10^(-3) molxx100gmol^(-1)`
`= 0.9375g = 0.94g`