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)) Whatmass 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.

Calcium carbonate reacts with aqueous HCl to give CaCl_(2) and CO_(2)

1.	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)) Whatmass 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.
Answer» Solution :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 with2 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`