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

`CaCO_(3)(s) + 2HCl(aq) to CaCl_(2)(aq) + CO_(2)(g) + H_(2)O(l)`
Mass of `CaCO_(3) = 1000g`
Moles of `CaCO_(3) = 1000/100 xx 10`mol
1000 mL of 0.76 M HCl contains HCl = 0.76 mol
250 mL of 0.76 M HCl contain HCl `=0.76/1000 xx 250`
=0.19 mol
Now, according to the equation, 1 mol of `CaCO_(3)`(s) requires 2 mol of HCl(aq). Hence, for 10 mol of `CaCO_(3)` (s), moles of HCl(aq) required would be
`=2/1 xx 10 = 20` mol
But we have only 0.19 mole of HCl (aq)
`therefore` HCl (aq) is limiting reagent.
2 mol of HCl (aq) from would from `CaCl_(2) =1/2 xx 0.19`
=0.095 mol
Mass of `CaCl_(2)` formed `=0.095 xx 111 = 10.54`g