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`

Explanation

In this problem, `CO_(2)` can be calculated by using either of the reactants `CaCO_(3)` or HCl. So, first of all we are to calculate the limiting reactant.
`CaCO_(3)(g) + 2HCl(aq) to CaCl_(2)(aq) + H_(2)O(l) + CO_(2)(g)`
According to above equation:
100 g of `CaCO_(3)` require 73 g of HCl
20 g of `CaCO_(3)` require `=73/100 xx 20 = 14.6`g
But amount of HCl actually present = 20.0 g Therefore, `CaCO_(3)` is limiting reactant and HCl is excess reactant.
Now let us calculate the amount of `CO_(2)` produced when entire quantity of limiting reactant reacts.
`underset(20 g)underset(100 g) underset(1 mol)(CaCO_(3)(s)) + 2HCl (g) to CaCl_(2)(aq) + H_(2)O(l) + underset(?)underset(44 g)underset(1 "mol")(CO_(2)(g))`
100 g of `CaCO_(3)` produces `CO_(3) = 44 g`
20 g of `CaCO_(3)` will produce `CO_(2) = 44/100 xx 20`
= 8.80 g