An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) together with other substances. Reaction of the ore with CO produces iron metal: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ (a) Balance this equation. (b) Calculate the number of grams of CO that can react with $0.350 \mathrm{~kg}\( of \)\mathrm{Fe}_{2} \mathrm{O}_{3}$ (c) Calculate the number of grams of Fe and the number of grams of \(\mathrm{CO}_{2}\) formed when \(0.350 \mathrm{~kg}\) of $\mathrm{Fe}_{2} \mathrm{O}_{3}$ reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

To balance the equation, we must ensure that there are equal numbers of each atom on both sides of the equation: Fe2O3(s) + CO(g) → Fe(s) + CO2(g) First, balance the Fe atoms: 2 Fe2O3(s) + CO(g) → 4 Fe(s) + CO2(g) Then balance the O atoms: 2 Fe2O3(s) + 3 CO(g) → 4 Fe(s) + 3 CO2(g) The equation is now balanced.

First, find the number of moles of Fe2O3 in 0.350 kg of material: Moles = (mass)/(molar mass) Fe molar mass = 55.845 g/mol O molar mass = 16.00 g/mol Molar mass of Fe2O3 = (2 × 55.85) + (3 × 16.00) = 159.69 g/mol Moles of Fe2O3 = (0.350 kg × 1000 g/kg) / 159.69 g/mol = 2.190 mol Now, use the balanced equation to find the moles of CO needed: 2 Fe2O3(s) + 3 CO(g) → 4 Fe(s) + 3 CO2(g) For every 2 moles of Fe2O3, we need 3 moles of CO Moles of CO = (2.190 mol Fe2O3) × (3 mol CO / 2 mol Fe2O3) = 3.285 mol Now, find the mass of these moles of CO: Molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol Mass of CO = (3.285 mol) × (28.01 g/mol) = 91.99 g So, 91.99 g of CO is needed to react with 0.350 kg of Fe2O3.

Use the balanced equation to find the moles of Fe formed: 2 Fe2O3(s) + 3 CO(g) → 4 Fe(s) + 3 CO2(g) For every 2 moles of Fe2O3, we get 4 moles of Fe Moles of Fe = (2.190 mol Fe2O3) × (4 mol Fe / 2 mol Fe2O3) = 4.380 mol Find the mass of these moles of Fe: Mass of Fe = (4.380 mol) × (55.845 g/mol) = 244.60 g of Fe Now, find the moles of CO2 formed: For every 2 moles of Fe2O3, we get 3 moles of CO2 Moles of CO2 = (2.190 mol Fe2O3) × (3 mol CO2 / 2 mol Fe2O3) = 3.285 mol Find the mass of these moles of CO2: Mass of CO2 = (3.285 mol) × (44.01 g/mol) = 144.51 g of CO2 So, 244.60 g of Fe and 144.51 g of CO2 are formed.

Total mass of reactants = Total mass of products Mass of Fe2O3 = 0.350 kg × 1000 g/kg = 350 g Mass of CO = 91.99 g Total mass of reactants = 350 g + 91.99 g = 441.99 g Mass of Fe = 244.60 g Mass of CO2 = 144.51 g Total mass of products = 244.60 g + 144.51 g = 389.11 g The mass of reactants (441.99 g) equals the mass of products (441.11 g). Therefore, the calculations are consistent with the law of conservation of mass.