Answer:
The reaction involved is:
\[\underset{2mol\,or\,54g}{\mathop{2Al+}}\,2NaOH+2{{H}_{2}}O\to
2NaAl{{O}_{2}}+\underset{3mol}{\mathop{3{{H}_{2}}}}\,\]
Number of
moles of H^ produced = \[\frac{3}{54}\times 0.15\]
\[=8333\times
{{10}^{-3}}\]
\[PV=nRT\]
\[1\times V=8.33\times
{{10}^{-3}}\times 0.083\times 293\]
\[V=0.202litre\]
\[=202mL.\]
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