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Q.
$100 \,g$ of $CaCO_3$ is treated with 1 litre of $1\, N\, HCl$. What would be the weight of $CO_2$ liberated after the completion of the reaction ?
Some Basic Concepts of Chemistry
Solution:
$1\, L $of $1N \,HCl = 1\, L$ of $1 \,M\, HCl = 1 \,mol \,of \,HC1$
Amount of $CaCO_{3} = 100\, g = 1 \,mol$
$\underset{1\,mol}{CaCO_3} + \underset {2\,mol}{2\,HCl} \to CaCl_2 + \underset {44\,g{(1\,mol)}}{CO_2 }+ H_2O$
Thus $HCl$ is the limiting reactant.
$2\,$ mol of $HCl$ after reaction with $CaCO_3$ give $CO_2= 1\, mol$
$1\, mol$ of $HCl$ after reaction with $CaCO_3$ give $CO_2 =$ mol.
$\therefore $ Mass of $CO_2$ obtained $= 1/2 \times 44 = 22\, g$