A) 22.4 L
B) 44.8 L
C) 5.6 L
D) 11.2 L
Correct Answer: C
Solution :
Eq of \[Al\]= eq of \[{{H}_{2}}\] \[\frac{4.5}{\frac{27}{3}}=\]eq of \[{{H}_{2}}\]; \[\frac{4.5}{9}=\]eq of \[{{H}_{2}}\] \[2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}\] eq. of \[{{H}_{2}}\] = Number of moles × n factor \[0.5=\,{{n}_{{{H}_{2}}}}\times 2\] \[{{V}_{{{H}_{2}}}}=\frac{0.5}{2}\times 22.4\]; \[{{V}_{{{H}_{2}}}}=5.6\ L\]
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