In the reaction, 2Al(s)+6HCl(aq)→2Al3+(aq.)+6Cl−(aq.)+3H2(g):
A
11.2H2(g) at STP is produced for every mole of HCl(ag.) consumed
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B
6LHCl(ag.) is consumed for every 3LH2(E) produced
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C
33.6LH2(g) is produced regardless of temperature and pressure for every mole of Al that reacts
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D
67.2LH2(g) at SIT is produced for every mac of Al that reacts
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Solution
The correct option is B11.2H2(g) at STP is produced for every mole of HCl(ag.) consumed So from the given reaction we can form 3×22.4 liters of H2 from 2 moles of HCl. So in the given reaction 11.2 liters of H2 is produced from 6 moles of HCl Hence option A is correct.