A piece of aluminium foil measuring 10.25 cm × 5.50 cm × 0.600 mm is dissolved in excess of HCl (aq). Density of foil is2.70 g/cm3. Thus, gaseous product formed is

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0.51 mol

1.52 mol

1.01 mol

3.00 mol

answer is A.

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Detailed Solution

2Al+6HCl⟶2AlCl3+3H2(g)2 mol 3 mol Volume of foil =10.25cm×5.50cm×0.60010cm= 3.3825 cm3Density :2.70 g/cm3∴ Mass of Al=3.3825×2.70=9.13275g Moles of Al=9.1327527=0.32825molH2 formed =1.5 times of Al reacted =1.5×0.33825=0.51mol

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