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

Calculation of mass of HCl
Molarity of solution `(M)=("Mass of HCl/Molar mass of HCl")/("Volume of solution in litres")`
`0.76 "mol L"^(-1)=(("Mass of HCl/36.5 g mol"))/((250//1000L))`
Mass of `HCl=(0.76"mol L"^(-1)xx(0.25L)(36.5"g mol"^(-1))=6.935g`
Step-II Identification of limiting reactant
`underset((100 g))(CaCO_(3)(s))+underset((2xx36.5=73g))(2HCl(aq))rarrunderset((111g))(CaCl_(2))(aq)+CO_(2)(g)+H_(2)O(l)`
100g of `CaCO_(3)` need HCl = 73 g
1000 g of `CaCO_(3)` need `HCl=((73g))/((100g))xx(1000g)=730g`
But the amount HCl actually available = 6.935 g
`:.` HCl is the limiting reactant
Step-III Calculation of mass/moles of `CaCl_(2)` in the reaction
73 g of HCl form `CaCl_(2) = 111g`
6.935 g of HCl form `CaCl_(2)=((111g))/((73g))xx(6.935g)=10.54g`
Moles of `CaCl_(2)` formed `= ("Mass of "CaCl_(2))/("Molar mass of CaCl"_(2))=((10.5g))/((111"g mol"^(-1)))=0.095` mol.