1.

If 20.0 g of CaCO_(3) is treated with 20.0 g of HCl, how many grams of CO_(2) will be produced ?

Answer»


Solution :`UNDERSET(40+12+3xx16=100g)(CaCO_(3))+underset(2(1+35.5)=73g)(2HCl)rarrCaCl_(2)+H_(2)O+underset(12+2xx16=44g.)(CO_(2))`
Here, `CaCO_(3)` will be the limiting reactant.


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