P_4O_6 reacts with water according to the equ | vedclass

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(B) Molecular mass of $P_4O_6 = 220 \ g/mol$. Moles of $P_4O_6$ dissolved = $\frac{1.1 \ g}{220 \ g/mol} = 0.005 \ mol$. According to the reaction $P_

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$400$

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(B) Molecular mass of $P_4O_6 = 220 \ g/mol$. Moles of $P_4O_6$ dissolved = $\frac{1.1 \ g}{220 \ g/mol} = 0.005 \ mol$. According to the reaction $P_4O_6 + 6H_2O 	o 4H_3PO_3$,$1 \ mol$ of $P_4O_6$ produces $4 \ mol$ of $H_3PO_3$. Moles of $H_3PO_3$ produced = $4 	imes 0.005 = 0.02 \ mol$. $H_3PO_3$ is a dibasic acid,so it reacts with $NaOH$ as: $H_3PO_3 + 2NaOH 	o Na_2HPO_3 + 2H_2O$. Moles of $NaOH$ required = $2 	imes 	ext{moles of } H_3PO_3 = 2 	imes 0.02 = 0.04 \ mol$. Volume of $NaOH$ solution = $\frac{	ext{moles}}{	ext{molarity}} = \frac{0.04 \ mol}{0.1 \ M} = 0.4 \ L = 400 \ mL$.

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

$P_4O_6$ reacts with water according to the equation $P_4O_6 + 6H_2O \to 4H_3PO_3$. Calculate the volume of $0.1 \ M$ $NaOH$ solution required to neutralize the acid formed by dissolving $1.1 \ g$ of $P_4O_6$ in $H_2O$. (in $mL$)

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