Question

A sample weighing 3.027 g is a mixture of Fe2O3 (molar mass = 159.69) and Al2O3 (molar mass = 101.96). When heat and a stream of H2 gas is applied to the sample, the Fe2O3 reacts to form metallic Fe and H2O(g). The Al2O3 does not react. If the sample residue (the solid species remaining after the reaction) weighs 2.889 g, what is the mass fraction of Fe2O3 in the original sample?

Answer 1

Fe2O3 + H2 = --> Fe(s) + H2O

balance:

Fe2O3 + 3/2H2 = --> 2Fe(s) + 3/2H2O

2Fe2O3 + 3H2 = --> 4Fe(s) + 3H2O

m final = 2.889 g (solids only)

find Fe2O3

first...

relate solids of Al2O3 + Fe(s)

mol of Fe2O3 = x

mol of Fe = y

mol of Al2O3 = z

so...

total mass of sampe 1 = 3.027 g

x*159.69 + z*101.96 = 3.027 (equation1)

for sample 2:

y*55.8 + z*101.96 = 2.889 (equation 2)

note that f we are talking about moles...

x = 2y (1 mol of Fe2O3 produces 2 mol of Fe(s))

x = 2y (equation 3)

substitute

2y *159.69 + z*101.96 = 3.027

y*55.8 + z*101.96 = 2.889

substract 1 from 2:

(2*159.69 - 55.8)y = 3.027 -2.889

y = ( 3.027 -2.889)/(2*159.69 - 55.8)

y = 0.0005235602 mol of Fe(s)

so..

x = 0.0005235602*2 = 0.0010471204 mol of Fe2O3

then

x*159.69 + z*101.96 = 3.027 (equation1)

0.0010471204 *159.69 + z*101.96 = 3.027 (equation1)

z= (3.027 -0.0010471204 *159.69 )/101.96

z = 0.02804811

mass of Fe2O3 = 0.0010471204 *159.69 = 0.16721

mass of Al2O3 = 0.02804811*101.96

total mass = 0.0010471204 *159.69 + 0.02804811*101.96 = 3.026999

% Fe2O3 = 0.16721 / 3.026999 * 100

% Fe2O3 = 5.5239 %