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Q.
The volume of carbon dioxide gas evolved at S.T.P. by heating $7.3\, g$ of $Mg(HCO_3)_2$ will be
Some Basic Concepts of Chemistry
Solution:
$Mg\left(HCO_{3}\right)_{2} \rightarrow MgO+2CO_{2}+H_{2}O$
$146\, g$ of $Mg\left(HCO_{3}\right)_{2}$ gives $22.4 × 2$ litre of $CO_{2}$
$\Rightarrow 7.3\,g of Mg\left(HCO_{3}\right)_{2}$ will give $\frac{22.4}{146}\times7.3\times2$
$2.24\,L$ of $CO_{2}=2240\,mL$ of $CO_{2}$
Since, there were 2 molecules of CO
thus = 1120 mL