To dissolve a mixture of oxides fe2o3 and feo weighing 10 g, 70 g of a solution of h2so4c W = 21%

To dissolve a mixture of oxides fe2o3 and feo weighing 10 g, 70 g of a solution of h2so4c W = 21% is required. Find the mass fraction of oxygen in a given mixture of oxides.

Given:
m mixture (Fe2O3, FeO) = 10 g m solution (H2SO4) = 70 g ω (H2SO4) = 21%
Find: ω in the mixture (О) -?
Decision:
1) write the reaction equations: Fe2O3 + 3H2SO4 => Fe2 (SO4) 3 + 3H2O FeO + H2SO4 => FeSO4 + H2O
2) find the mass of dry matter MeOH: m (H2SO4) = m solution (H2SO4) * ω (H2SO4) / 100% = 70 * 21% / 100% = 14.7 g
3) find the total amount of substance H2SO4: n total. (H2SO4) = m (H2SO4) / Mr (H2SO4) = 14.7 / 98 = 0.15 mol
4) take the amount of the substance Fe2O3 for X, then the amount of the substance H2SO4 that reacted with Fe2O3 will be 3x (according to the reaction equation) n (Fe2O3) = x mol n1 (H2SO4) = 3 * n (Fe2O3) = 3x mol
5) find the amount of substance H2SO4 reacted with FeO: n2 (H2SO4) = n total. (H2SO4) – n1 (H2SO4) = 0.15 – 3x mol
6) find the amount of the substance FeO (according to the reaction equation): n (FeO) = n2 (H2SO4) = 0.15 – 3x mol
7) compose and solve the equation for finding the mass of a mixture of Fe2O3 and FeO: m mixture (Fe2O3, FeO) = m (Fe2O3) + m (FeO) = n (Fe2O3) * Mr (Fe2O3) + n (FeO) * Mr (FeO ) 10 = x * 160 + (0.15 – 3x) * 72 160x + 10.8 – 216x = 10 56x = 0.8 x = 0.014
8) find the amount of the substance Fe2O3 and FeO: n (Fe2O3) = x = 0.014 mol n (FeO) = 0.15 – 3x = 0.15 – 3 * 0.014 = 0.108 mol
9) find n (O) in a mixture of Fe2O3 and FeO: n (O in Fe2O3) = 3 * n (Fe2O3) = 3 * 0.014 = 0.042 mol n (O in FeO) = n (FeO) = 0.108 mol n (O ) = n (O in Fe2O3) + n (O in FeO) = 0.042 + 0.108 = 0.15 mol
10) find the mass of O: m (O) = n (O) * Mr (O) = 0.15 * 16 = 2.4 g 11) find the mass fraction of O in the mixture: ω in the mixture (O) = m (O ) * 100% / m mixture (Fe2O3, FeO) = 2.4 * 100% / 10 = 24%
Answer: the mass fraction of oxygen (O) in the mixture (Fe2O3, FeO) is 24%.