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

Answer

`0.94 g Ca Co_(3)`

Explanation

0.75 M of HCl `-=` 0.75 mol of HCl are present in 1 L water
`-=[(0.75 mol)xx(36.5 g mol^(-1))]` HCl is present in 1 L of water
`-=` 27.375 g of HCl is Present in 1 L of Water
Thus ,1000 mL of solution sontains 27.375 g of HCl.
`therefore` Amount of HCl present in 25 mL of solution
`=(27.375 g)/(1000 mL)xx25 ml`
=0.6844 g
From the given chemical equation ,
`CaCO_(3(s))+2HCl_(aq) to CaCl_(2(aq))+CO_(2(g))+H_(2)O_((l))`
2 mol of HCl `(2xx36.5 =71 g)` react with 1 mol of`CaCO_(3)` (100 g)
`therefore`Amount of `CaCo_(3)` that will react with 0.6844 g` (100)/(71)xx0.6844g`
=0.9639 g