Calculate the approximate mass (in g) of H_2S | vedclass

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(B) The balanced chemical equation is:$2 H_2S_{(g)} + 3 O_{2(g)} \longrightarrow 2 SO_{2(g)} + 2 H_2O_{(g)}$From the stoichiometry, $3 \ mol$ of $O_2$

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$15.16$

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(B) The balanced chemical equation is:$2 H_2S_{(g)} + 3 O_{2(g)} \longrightarrow 2 SO_{2(g)} + 2 H_2O_{(g)}$From the stoichiometry, $3 \ mol$ of $O_2$ reacts with $2 \ mol$ of $H_2S$.At $STP$, $1 \ mol$ of gas occupies $22.4 \ L$.Therefore, $15 \ L$ of $O_2 = \frac{15}{22.4} \ mol \approx 0.6696 \ mol$.Using the mole ratio, the moles of $H_2S$ required = $\frac{2}{3} 	imes \left(\frac{15}{22.4}ight) \ mol$.Mass of $H_2S = 	ext{moles} 	imes 	ext{molar mass} = \left(\frac{2}{3} 	imes \frac{15}{22.4}ight) 	imes 34 \ g \ mol^{-1}$.Mass of $H_2S = \frac{10}{22.4} 	imes 34 \approx 15.178 \ g$.The approximate mass is $15.16 \ g$.

Calculate the approximate mass (in $g$) of $H_2S$ required for the following reaction when $15 \ L$ of oxygen at $STP$ reacts completely.
$2 H_2S_{(g)} + 3 O_{2(g)} \longrightarrow 2 SO_{2(g)} + 2 H_2O_{(g)}$
(molar mass of $H_2S = 34 \ g \ mol^{-1}$)

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Calculate the approximate mass (in $g$) of $H_2S$ required for the following reaction when $15 \ L$ of oxygen at $STP$ reacts completely.$2 H_2S_{(g)} + 3 O_{2(g)} \longrightarrow 2 SO_{2(g)} + 2 H_2O_{(g)}$(molar mass of $H_2S = 34 \ g \ mol^{-1}$)