A) 44.8 L
B) 22.4 L
C) 11.2L
D) 5.6 L
Correct Answer: D
Solution :
\[2{{H}^{+}}+2{{e}^{-}}\xrightarrow{{}}{{H}_{2}}\] \[A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}\,Al\] \[\frac{w({{H}_{2}})}{w(A)}=\frac{E({{H}_{2}})}{E\,(Al)}\] \[=\frac{1}{9}\] \[w({{H}_{2}})=4.5\times \frac{1}{9}\] \[=0.5\,\,g\] \[=\frac{1}{4}\,\,mol\,{{H}_{2}}\] \[=\frac{22.4}{4}=5.6\,L\,{{H}_{2}}\] at \[STP\]You need to login to perform this action.
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