A mixture of aluminum and iron weighing 9.62 g reacts with hydrogen chloride in aqueous solution according to the parallel reactions 2 Al + 6 HCl - > 2 AlCl3 + 3 H2 Fe + 2 HCl - > FeCl2 + H2 A 0.738 g quantity of hydrogen is evolved when the metals react completely. Calculate the mass of iron in the original mixture. Suppose that in the previous reaction I also get 15.95 g AlCl3. What is the percent yield of AlCl3?

for a)

mAl = 3.87 gr

mFe = 5.75 gr

for b)

Y = 13.1%

Explanation:

since aluminium reacts according to

2 Al + 6 HCl - > 2 AlCl₃ + 3 H₂

then 2 moles of Al generates 3 moles of H2

then the mass of hydrogen gas obtained per mass of aluminium that reacts is

R₁ = 3 n H₂ / 2 n Al = [3 mol * 2 gr/mol] / [2 mol * 27 gr/mol ] = 1/9

then for iron

Fe + 2 HCl - > FeCl₂ + H₂

the mass of hydrogen gas obtained per mass of iron # that reacts is

R₂ = 3 n H₂ / 2 n Fe = [3 mol * 2 gr/mol] / [2 mol * 56 gr/mol ] = 3/56

then the mass of hydrogen obtained is

mAl*R₁ + mFe*R₂ = mH

mAl+mFe = mM

where m represents mass, M the total mass of the metals, H the mass of hydrogen

thus

mAl = mM - mFe

and

mAl*R₁ + (mM-mAl) * R₂ = mH

mAl * (R₁-R₂) + mM * R₂ = mH

mAl = (mH - mM * R₂) / (R₁-R₂)

replacing values

mAl = (mH - mM * R₂) / (R₁-R₂) = (0.738 g - 9.62 g*3/56) / (1/9-3/56) = 3.87 gr

and

mFe = mM - mAl = 9.62 g - 3.87 gr = 5.75 gr

for b)

since

m AlCl3 = m AlCl3 * (m Fe / mAlCl3) = 33.87 g * (97 g / 27 g) = 121.68 g

the yield of AlCl3 is

Y = 15.95 g / 121.68 g = 0.131 = 13.1%