What volume of hydrogen gas at STP would be l | vedclass

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(C) The chemical equation for the reaction is: $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$.First, calculate the mass of pure $H_2SO_4$:Mass of solution $=

Vedclass · Accepted Answer

$6.892$

Vedclass · Answer

(C) The chemical equation for the reaction is: $Zn + H_2SO_4 ightarrow ZnSO_4 + H_2$.First, calculate the mass of pure $H_2SO_4$:Mass of solution $= 	ext{Volume} 	imes 	ext{Density} = 50 	ext{ mL} 	imes 1.3 	ext{ g mL}^{-1} = 65 	ext{ g}$.Mass of pure $H_2SO_4 = 65 	ext{ g} 	imes 0.50 = 32.5 	ext{ g}$.Moles of $H_2SO_4 = \frac{32.5 	ext{ g}}{98 	ext{ g mol}^{-1}} \approx 0.3316 	ext{ mol}$.Moles of $Zn = \frac{20 	ext{ g}}{65 	ext{ g mol}^{-1}} \approx 0.3077 	ext{ mol}$.Since the stoichiometry of the reaction is $1:1$, $Zn$ is the limiting reagent because it has fewer moles.Therefore, moles of $H_2$ produced $= 	ext{moles of } Zn = 0.3077 	ext{ mol}$.Volume of $H_2$ at $STP = 0.3077 	ext{ mol} 	imes 22.4 	ext{ L mol}^{-1} \approx 6.892 	ext{ L}$.

What volume of hydrogen gas at $STP$ would be liberated by the action of $50 \text{ mL}$ of $H_2SO_4$ of $50\%$ purity (density = $1.3 \text{ g mL}^{-1}$) on $20 \text{ g}$ of zinc (in $L$)? Given: Molar mass of $H, O, S, Zn$ are $1, 16, 32, 65 \text{ g mol}^{-1}$ respectively.

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