Calcium carbonate reacts with aqueous HCl to give `CaCl_(2)` and `CO_(2)` according to the reaction, `CaCO_(3) (s) + 2 HCl (aq) to CaCl_(2) (aq) + CO_(2)(g) + H2O(l) `
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Explanation

Molarity (M) of a solution is given by:
`w=(MM.V)/1000`
In the present case, M = 0.75, M. = 1 + 35.5 = 36.5, V = 25 mL, w =? Substituting the values, we have
`w=(0.75 xx 36.5 xx 25)/1000 = 0.6844 g`
Therefore, the mass of HCI present in the given solution is 0.6844 g.
The given equation is:
`underset("1 mole 100 g")(CaCO_(3)(s)) + underset("2 moles " 2 xx 36.5 = 73 g)(2HCl(aq)) to CaCl_(2)(aq) + CO_(2)(g) + H_(2)(l)`
`therefore` 73 g of HCI require for complete reaction the mass of `CaCO_3 = 100 g`
`therefore` 0.6844 g of HCI will require `CaCO_(3) = 100/73 xx 0.6844 = 0.939 g`