Answer:
The given reaction is:
\[\begin{align}
&
{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}O(g) \\
&
\begin{matrix}
\,\,\,1\,mol\,\,\,\, & \,\,\,\,\,\,\,\frac{1}{2}mol &
\,\,\,\,\,\,\,\,\,\,\,\,1\,mol \\
\end{matrix}
\\
&
\begin{matrix}
\,\,\,1\,vol\,\,\,\, & \,\,\,\,\,\,\,\frac{1}{2}vol &
\,\,\,\,\,\,\,\,\,\,\,\,1\,vol \\
\end{matrix}
\\
&
\begin{matrix}
\,\,\,10\,vol\,\, & \,\,\,\,\,\,\,5\,vol &
\,\,\,\,\,\,\,\,\,\,\,\,10\,vol \\
\end{matrix}
\\
\end{align}\]
Thus, 10 volumes of water vapour will be obtained.
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