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In the reaction $2Al(s)+6HCl(aq)\rightarrow 2Al^{3+}(aq)+6Cl^{-1}(aq)+3H_2(g)$

$\begin{array}{1 1}(a)\;6\;litre\;HCl(aq)\;is\;consumed\;for\;every\;3LH_2(g)\;produced\\(b)\;33.6litre\;H_2(g)\;is\;produced\;regardless\;of\;temperature\;and\;pressure\;for\;every\;mole\;that\;reacts\\(c)\;67.2litre \;H_2(g)\;at\;STP\;is\;produced\;every\;mole\;Al\;that\;reacts\\(d)\;11.2litre\;H_2(g)\;at\;STP\;is\; produced\;every\;mole\;HCl(aq)\;consumed\end{array}$

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$2Al(s)+6HCl(aq)\rightarrow 2Al^{3+}(aq)+6cl^{-1}(aq)$
For each mole of HCl reacted ,0.5mole of $H_2$ gas is formed at STP.
1 mole of an ideal gas occupies $22.4l$ STP.
Volume of $H_2$ gas formed at STP per mole of HCL reacted is $22.4\times 0.5=11.2l$
Hence (d) is the correct answer.
answered Oct 30, 2013 by sreemathi.v

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