A) 22.4 L
B) 44.8 L
C) 5.6 L
D) 11.2 L
Correct Answer: C
Solution :
[c] From second law of Faraday |
\[\frac{{{m}_{Al}}}{{{m}_{H}}}=\frac{{{E}_{Al}}}{{{E}_{H}}}\] |
\[\therefore \frac{4.5}{{{m}_{H}}}=\frac{27/3}{1}\] |
\[or{{m}_{H}}=0.5\,g\] |
\[\because \,2\,g\,{{H}_{2}}\,volume\,at\,STP=22.4\,L\] |
\[\therefore \,\,0.5\,g\,{{H}_{2}}\,volume\,\,at\,STP\] |
\[=\frac{22.4\times 0.5}{2}\,L=5.6\,L\] |
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