Calcium carbonate reacts with aqueous HCl to give CaCl_(2) and CO_(2) according to the reaction : 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 react 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 : 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 react with 1000 g of CaCO_(3)? Name the limiting reagent. Calculate the number of moles of CaCl_(2) formed in the reaction.
Answer» Solution :`"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))`