What volume (in mL) of HCl solution containin | vedclass

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(C) The reaction of metallic sodium with water is: $2Na + 2H_2O \rightarrow 2NaOH + H_2$. Moles of $Na = \frac{0.46 \ g}{23 \ g/mol} = 0.02 \ mol$. Si

Vedclass · Accepted Answer

$10$

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(C) The reaction of metallic sodium with water is: $2Na + 2H_2O ightarrow 2NaOH + H_2$. Moles of $Na = \frac{0.46 \ g}{23 \ g/mol} = 0.02 \ mol$. Since $2 \ mol$ of $Na$ produce $2 \ mol$ of $NaOH$,$0.02 \ mol$ of $Na$ will produce $0.02 \ mol$ of $NaOH$. The neutralization reaction is: $NaOH + HCl ightarrow NaCl + H_2O$. To neutralize $0.02 \ mol$ of $NaOH$,we require $0.02 \ mol$ of $HCl$. Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$. Concentration of $HCl$ solution = $73 \ g/L = \frac{73 \ g/L}{36.5 \ g/mol} = 2 \ M$. Volume of $HCl$ required = $\frac{	ext{moles}}{	ext{molarity}} = \frac{0.02 \ mol}{2 \ mol/L} = 0.01 \ L = 10 \ mL$.

List-$I$	List-$II$
$A. 10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$1. 0.224 \ L \ CO_2$
$B. 1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl} \text{reaction}$	$2. 4.48 \ L \ CO_2$
$C. 2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$3. 0.448 \ L \ CO_2$
$D. 0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$4. 2.24 \ L \ CO_2$
	$5. 22.4 \ L \ CO_2$

What volume (in $mL$) of $HCl$ solution containing $73 \ g$ per litre is required to completely neutralise sodium hydroxide solution,obtained by allowing $0.46 \ g$ of metallic sodium to act upon water?

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