Al(s) + HCl(aq) AlCl3(aq) + H2(g) Consider the above unbalanced equation. A 1.00 g sample of a copper-aluminum alloy is reacted with excess hydrochloric acid. Only the aluminum reacts. The H2 produced is collected by displacement of water. If 110. mL of H2 is collected at a barometric pressure of 725 mm Hg and 25.0°C, what is the mass percent of Al in the sample? The vapor pressure of water is 23.8 mm Hg at 25.0°C. Use molar masses with at least as many significant figures as the data given.
Pressure of hydrogen = P(total) - P(water)
= 725 - 23.8
= 701.2 mmHg or 0.923 atm
Temperature = 25°C = 298 K
Moles of Hydrogen = pressure * Volume / ( gas constant * temperature)
= ( 0.923 * 0.110 ( in litre )) / ( 0.082 * 298 )
= 4.155*10^-3 moles
2 Al + 6 HCl =2 AlCl3 + 3 H2
Moles Al = 2/3 * moles of H2
= 2/3 * ( 4.155*10^-3)
= 2.77*10^-3
Mass of Al = mole * molar mass
= ( 2.77*10^-3) * 27
= 0.0748 grams
Mass % of aluminium = mass of aluminium / total mass ) *100
= 0.0748 / 1 )*100
= 7.48%
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