What amount of oxygen is used at S.T.P to obt | vedclass

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(A) The balanced chemical equation for the reaction is: $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$From the stoichiometry,$1 \ 	ext{

Vedclass · Accepted Answer

$5.6$

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(A) The balanced chemical equation for the reaction is: $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$From the stoichiometry,$1 \ 	ext{mole}$ of $H_2O$ $(18 \ g)$ is produced by $0.5 \ 	ext{moles}$ of $O_2$.At $S.T.P$,$1 \ 	ext{mole}$ of gas occupies $22.4 \ dm^3$,so $0.5 \ 	ext{moles}$ of $O_2$ occupies $11.2 \ dm^3$.Thus,$11.2 \ dm^3$ of $O_2$ is required to produce $18 \ g$ of $H_2O$.For $9 \ g$ of $H_2O$,the volume of $O_2$ required is: $\frac{11.2 \ dm^3}{18 \ g} 	imes 9 \ g = 5.6 \ dm^3$.

What amount of oxygen is used at $S.T.P$ to obtain $9 \ g$ water from a sufficient amount of hydrogen gas (in $dm^3$)?

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