Calculate the volume and the number of molecules of `CO_2` liberated at STP if 50 grams of `CaCO_3` is treated with dilute hydrochloric acid which contains 7.3 grams of dissolved HCl gas.

Explanation

As per the metric staichio equation 100g of `CaCO_3` reacts with 73g of HCl to liberate 44 g of `CO_2`
In the above problem the amount of `CaCO_3` taken is 50g and HCl availble is 7.3g.
100g of `CaCO,_3`require 73g of HCl and 50g of `CaCO_3` required 36.5g of HCl but, only 7.3g of HCl is available.
Hence the product `CO_2` formed depends only on the amount of HCl which is in the least amount but not on the amount of `CaCO_3` which is an excess. The reactant available in less amount is called limiting reagent as it limits the amount of product formed.
therefore, we can write
73 g of HCl `rarr` 44 g of `CO_2`
7.3 g of HCl - ?
`(7.3g xx 44g)/(73g}` = 4.4 g
44 g of `CO_2` occupies 22.4 L volume at STP
4.4 g of `CO_2` occupies - ?
`(4.4g xx" 22.4 litres")/(44 g)` = 2.24 litres
44 g of `CO_2` contain `6.023xx10^(23)` mol of `CO_2`
4.4 g Contains - ?
`(4.4g xx 6.023 xx 10^(23))/(44 g) = 6.023 xx 10^(22)` mol