A) 44.8 L
B) 11.2 L
C) 22.4 L
D) 5.6 L
Correct Answer: D
Solution :
Equivalent mass of\[Al=\frac{27}{3}=9\] Equivalent mass of \[H=1\] \[\frac{{{W}_{Al}}}{{{W}_{{{H}_{2}}}}}=\frac{eq.\,mass\,of\,Al}{eq.\,mass\,of\,{{H}_{2}}}\] \[\frac{4.5}{{{W}_{{{H}_{2}}}}}=\frac{9}{1}\] \[{{W}_{{{H}_{2}}}}=0.5\,g\] \[\because \]2 g of\[{{H}_{2}}\]at STP occupy volume =22.4L \[\therefore \]0.5 g of\[{{H}_{2}}\]at STP will occupy volume \[=\frac{22.4\times 0.5}{2}\] \[=5.6L\]You need to login to perform this action.
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