What will be the amount of MgCl_2 produced wh | vedclass

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(B) The balanced chemical equation is: $MgO + 2HCl \longrightarrow MgCl_2 + H_2O$ From the stoichiometry of the reaction,$2 \ mol$ of $HCl$ reacts to

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$22.12$

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(B) The balanced chemical equation is: $MgO + 2HCl \longrightarrow MgCl_2 + H_2O$ From the stoichiometry of the reaction,$2 \ mol$ of $HCl$ reacts to produce $1 \ mol$ of $MgCl_2$. Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$. Mass of $2 \ mol$ of $HCl = 2 	imes 36.5 = 73 \ g$. Molar mass of $MgCl_2 = 24 + 2 	imes 35.5 = 95 \ g/mol$. According to the reaction,$73 \ g$ of $HCl$ produces $95 \ g$ of $MgCl_2$. Therefore,$17 \ g$ of $HCl$ will produce: $\frac{95 \ g}{73 \ g} 	imes 17 \ g \approx 22.12 \ g$ of $MgCl_2$.

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

What will be the amount of $MgCl_2$ produced when $17 \ g$ of $HCl$ is reacted with an excess of $MgO$ in the following reaction?
$MgO + 2HCl \longrightarrow MgCl_2 + H_2O$

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What will be the amount of $MgCl_2$ produced when $17 \ g$ of $HCl$ is reacted with an excess of $MgO$ in the following reaction? $MgO + 2HCl \longrightarrow MgCl_2 + H_2O$