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