Calcium carbonate reacts with aqueous HCl to give `CaCl_(2)` and `CO_(2)` according to the reaction given below
`CaCO_(3)(s)+2HCl(aq)rarrCaCl_(2)(aq)+CO_(2)(g)+H_(2)O(l)`
What mass of `CaCl_(2)` will be formed when 250mL of 0.76 M HCl reac ts with 1000 g of `CaCO_(3)`? Name the limiting reagent. Calculate the number of moles of `CaCl_(2)` formed in the reaction.

Explanation

Number of moles of HCl = `250 mL xx (0.76 M)/(1000) 0.19` mol
Mass of `CaCO_(3) = 1000 g` Number of moles `CaCO_(3) = (1000 g)/(100 g) = 10` mol
According to given equation 1 mol of `CaCO_(3)` (s) requires 2 mol of HCl (aq). Hence, for the reaction of 10 mol of `CaCO_(3)` (s) number of moles of HCl required would be:
10 mol `CaCO_(3) xx ("2 mol HCl(aq)")/("1 mol " CaCO_(3)) = "20 mol HCl(aq)"`
But we have only 0.19 mol HCl (aq), hence, HCl (aq) is limiting reagent.
So amount of `CaCl_(2)` formed will depend on the amount of HCl available.
Since, 2 mol HCl (aq) forms 1 mol of `CaCl_(2)`, therefore, 0.19 mol of HCl (aq) would give:
`0.19 "mol HCl (aq) "xx ("1 mol of " CaCl_(2)(aq))/("2mol HCl(aq)") = 0.095 mol` or `0.095 xx "molar mass of " CaCl_(2) = 0.095 xx 111 = 10.54g`