Question

If 20g of `CaCO_(3)` is treated with 20g of HCl. How many grams of `CO_(2)` can be generated according to the following equations?
`CaCO_(3)+2HCl(aq) to CaCl_(2)(aq.)+H_(2)O(l)+CO_(2)g`

Answer 1

`underset(100g)underset("1 mol")(CaCO_(3))+underset(73g)underset("2 mol")(2HCl(aq)) to CaCl_(2)(aq.)+H_(2)O(l)+underset(44g)underset("1 mol")(CO_(2)g)`
Let `CaCO_(3)(s)` be completely consumed in the reaction.
`therefore 100g CaCO_(3) "give" 44gCO_(2)`
`therefore 20g CaCO_(3)"will give" (44)/(100)xx20g CO_(2)=8.8CO_(2)`
Let HCl be completely consumed.
`therefore 73g "HCl give" 44g CO_(2)`
`therefore 20 "g of HCl will give" (44)/(73)xx20 g CO_(2)=12.054g CO_(2)`
Since, `CaCO_(3)` gives least amount of product `CO_(2)`, hence, `CaCO_(3)` is limiting reactant. Amount of `CO_(2)` formed will be 8.8g