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

Explanation

Molar mass of `CaCO_(3)=40+12+3xx16`
`=100 g "mol"^(-1)`
Moles of `CaCO_(3)` in 1000 g
`n_(CaCO_(3))=("Mass")/("Molar mass")`
`n_(CaCO_(3))=(1000)/(100)=10 `mol
Molarity `=("Moles of solute"xx 1000)/("Volume of solution")`
`0.76=(n_(HCl)xx1000)/(250)`
`n_(HCl)=(0.76xx250)/(1000)=0.19` mol
`{:(CaCO_(3(s)),+,2HCl_((aq)),rarr,CaCl_(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
`:.` 10 moles of `CaCO_(3)` will react with `(10xx2)/(1)=20` moles of 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 `CaCO_(2)` 0.19 mole of HCl will produce `(1xx0.19)/(2)=0.095`
mol of `CaCl_(2)`
Molar mass of `CaCl_(2)=40+(2xx3.35.5)=111g "mol"^(-1)`
`:.0.095` mole of `CaCl_(2)=0.095xx111=10.54g`