A) \[112L{{H}_{2(g)}}\]at STP is produced for every mole \[HC{{l}_{(aq)}}\] consumed
B) \[6L\,HC{{l}_{(aq)}}\]is consumed for every 2 L \[{{H}_{2(g)}}\]produced
C) \[33.6L\,{{H}_{2(g)}}\]is produced regardless of temperature and pressure for every mole Al that reacts
D) \[67.2L{{H}_{2(g)}}\]at STP is produced for every mole Al that reacts.
Correct Answer: C
Solution :
[c]:\[2A{{l}_{(s)}}+6HC{{l}_{(aq)}}\xrightarrow[{}]{{}}\] \[2Al_{(aq)}^{3+}+6Cl_{(aq)}^{-}+3{{H}_{2(g)}}\] At STP, 6 moles of HCl produces 3 moles of \[{{H}_{2}}\] or \[3\times 22.4\]lit of \[{{H}_{2}}\] \[\therefore \]1 mole of HCl produces \[\frac{3\times 22.4}{6}=11.2\]lit of \[{{H}_{2}}\] Again at STP, 2 moles of Al produces 3 moles of\[{{H}_{2}}\]or \[3\times 22.4\]lit of \[{{H}_{2}}\]or 1 mole of Al produces \[\frac{3\times 22.4}{2}=33.6\]lit of \[{{H}_{2}}\]
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