Question

A sample weighing 3.099 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 A1203 does not react. If the sample residue (the solid species remaining after the reaction) weighs 2.498 g, what is the mass fraction of Fe2O3 in the original sample? Number g FeO g original sample

Answer 1

Fe2O3 + Al2O3 +3 H2 (g) $\rightarrow$ 2Fe+Al2O3 + 3H2O

weight of ( Fe2O3 + Al2O3) before reaction =3.099g

weight of solid residue after reaction (Fe+Al2O3) =2.498g

Diferrence=3.099-2.498=0.601g weight of water vapours evolved

moles of water vapours=0.601g/18g/mol=0.033moles of H2O

But 3 moles of H2O(g) = 1moles of Fe2O3 (stoichometric coefficients)

so 0.033 moles H2O (g)=1/3* 0.033=0.011 moles of Fe2O3(s)

                                                   =0.011moles *159.69 g/mol

                                                  =1.773 g of Fe2O3(s)

so mass fraction of Fe2O3(s)=1.773/3.099=0.57

Answer 2

Fe2O3 + Al2O3 +3 H2 (g) $\rightarrow$ 2Fe+Al2O3 + 3H2O

weight of ( Fe2O3 + Al2O3) before reaction =3.099g

weight of solid residue after reaction (Fe+Al2O3) =2.498g

Diferrence=3.099-2.498=0.601g weight of water vapours evolved

moles of water vapours=0.601g/18g/mol=0.033moles of H2O

But 3 moles of H2O(g) = 1moles of Fe2O3 (stoichometric coefficients)

so 0.033 moles H2O (g)=1/3* 0.033=0.011 moles of Fe2O3(s)

                                                   =0.011moles *159.69 g/mol

                                                  =1.773 g of Fe2O3(s)

so mass fraction of Fe2O3(s)=1.773/3.099=0.57

Answer 3

Fe2O3 + Al2O3 +3 H2 (g) $\rightarrow$ 2Fe+Al2O3 + 3H2O

weight of ( Fe2O3 + Al2O3) before reaction =3.099g

weight of solid residue after reaction (Fe+Al2O3) =2.498g

Diferrence=3.099-2.498=0.601g weight of water vapours evolved

moles of water vapours=0.601g/18g/mol=0.033moles of H2O

But 3 moles of H2O(g) = 1moles of Fe2O3 (stoichometric coefficients)

so 0.033 moles H2O (g)=1/3* 0.033=0.011 moles of Fe2O3(s)

                                                   =0.011moles *159.69 g/mol

                                                  =1.773 g of Fe2O3(s)

so mass fraction of Fe2O3(s)=1.773/3.099=0.57