Amixture 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%

14 days is the half-life of phosphorus-32