0.4 text{ g} of dihydrogen is made to react w | vedclass

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(B) The balanced chemical equation is: $H_{2(g)} + Cl_{2(g)} ightarrow 2HCl_{(g)}$Moles of $H_2 = \frac{0.4 	ext{ g}}{2 	ext{ g/mol}} = 0.2 	ext{

Vedclass · Accepted Answer

$4.67$

Vedclass · Answer

(B) The balanced chemical equation is: $H_{2(g)} + Cl_{2(g)} ightarrow 2HCl_{(g)}$Moles of $H_2 = \frac{0.4 	ext{ g}}{2 	ext{ g/mol}} = 0.2 	ext{ mol}$Moles of $Cl_2 = \frac{7.4 	ext{ g}}{71 	ext{ g/mol}} \approx 0.1042 	ext{ mol}$Since $1 	ext{ mol}$ of $H_2$ reacts with $1 	ext{ mol}$ of $Cl_2$, $Cl_2$ is the limiting reagent because it is present in a smaller amount.Moles of $HCl$ formed $= 2 	imes 	ext{moles of } Cl_2 = 2 	imes 0.1042 = 0.2084 	ext{ mol}$At $273 	ext{ K}$ and $1 	ext{ bar}$ ($STP$ conditions), the molar volume of an ideal gas is approximately $22.7 	ext{ L/mol}$.Volume of $HCl = 0.2084 	ext{ mol} 	imes 22.7 	ext{ L/mol} \approx 4.73 	ext{ L}$.Given the options, $4.67 	ext{ L}$ is the closest value, which is obtained by using the older $STP$ definition $(22.4 	ext{ L/mol})$.Thus, the correct option is $(b)$.

$0.4 \text{ g}$ of dihydrogen is made to react with $7.4 \text{ g}$ of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at $273 \text{ K}$ and $1 \text{ bar}$ pressure is: (in $\text{ L}$)

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