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

`"1000 g of CaCO"_(3)="1000/100 mol = 10 mol "("Molecular mass of "CaCO_(3)="100 g mol"^(-1))`
`"250 mL of 0.76 M HCl "=250xx0.76" millimole = 190 millimole = 0.19 mol"`
According to the given equation, 1 mol of `CaCO_(3)` reacts with 2 mol of HCl
`therefore" 10 mol of "CaCO_(3)" will react with 20 mol of HCl"`
But we have only 0.19 mol HCl. Hence, HCl will be limiting reagent.
2 mol of HCl produce `CaCl_(2)` = 1 mol
`therefore " 0.19 mol of HCl will produce "CaCl_(2)=(1)/(2)xx0.19" mol"=" 0.095 mol"= 0.095xx111g=10.54h`
`" "("Molar mass of "CaCl_(2)=40+71="111 g mol"^(-1))`