In the reaction. 2Al(s)+6HCl(s)→2Al(aq) 3++6Cl(aq) −+3H2(g)
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a
6 lit HCl(aq) is consumed for every 3 lit H2(g)
b
33.6lit H2(aq) is produced regardless of temperatureand pressure for every mole Al that reacts
c
67.2 lit H2(g) at STP is produced for every mole Al that reacts
d
11.2 lit H2(aq) is produced for every mole HCl(aq) consumed
answer is D.
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Detailed Solution
2Al(s)+6HCl(aq)⟶2Al3+(aq)+6Cl−(aq)+3H2(g)For each mole of HCl reacted, 0.5 mole of H2 gas is formed at STP. 1 mole of an ideal gas occupies 22.4 lit at STP. Volume of H2 gas formed at STP per mole of HClreacted is 22.4×0.5 litre.