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