Question

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.

Answer 1

Molar mass of `CaCO_(3)=40+12+3xx16=100g "mol"^(-1)`
Moles of `CaCO_(3)` in `1000g,n_(CaCO_(3))=("mass"(g))/("molar mass")`
`n_(CaCO_(3))=(1000g)/(100g "mol"^(-1))=10` mol
Molarity `=("moles of solute"(HCl)xx1000)/("Volume of solution")`
(It is giv en that moles of HCl in 250mL of 0.76M HCl`=n_(HCl)`)
`0.76=(n_(HCl)xx1000)/(250)`
`n_(HCl)=(0.76xx250)/(1000)=0.19` mol
`CaCO_(3)(s)+2HCl(Aq)rarrCaCl_(2)(aq)+CO_(2)(g)+H_(2)O(l)`
1 mol 2 mol
According to the equation
1 mole of `CaCO_(3)` reacts with 2 moles HCl
`therefore` 10 moles of `CaCO_(3)` will reacts with 2 moles HCl
But we have only 0.19 moles HCl, so HCl is limiting reagent and it limits the yield of `CaCl_(2)`.
Since, 2 moles of HCl produces 1 mole of `CaCl_(2)`
Molar mass of `CaCl_(2)=40+(2xx35.5)=111g "mol"^(-1)`
`therefore 0.095 "mole of" CaCl_(2)=0.095xx111=10.54g`