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 $$ 0.350 \mathrm{~kg} \text { of } \mathrm{Fe}_{2} \mathrm{O}_{3} $$ (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.

Step by step solution

01

1. Balancing the chemical equation

To balance the chemical equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. The unbalanced equation is: Fe₂O₃ (s) + CO (g) → Fe (s) + CO₂ (g) To balance the equation, we can start by balancing the Fe atoms \(2 \text{Fe}_2 O_3 (s) + CO (g) \longrightarrow 4 Fe (s) + CO_2 (g)\) Now, we can balance the O atoms: \(Fe_2 O_3 (s) + 3CO (g) \longrightarrow 2Fe (s) + 3CO_2 (g)\) The balanced equation is: \(Fe_2 O_3 (s) + 3CO (g) \longrightarrow 2Fe (s) + 3CO_2 (g)\)

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2. Calculate the grams of CO that can react with 0.350 kg of Fe2O3

First, we need to convert the mass of Fe₂O₃ from kg to g: 0.350 kg Fe₂O₃ × (1000 g / 1 kg) = 350 g Fe₂O₃ Next, we find the molar mass of Fe₂O₃: \(M_{Fe_2 O_3} = 2(55.845 \text{g/mol}) + 3(16.00 \text{g/mol}) = 159.69 \text{g/mol}\) Now, find the moles of Fe₂O₃: moles of Fe₂O₃ = (mass of Fe₂O₃) / (molar mass of Fe₂O₃) = (350 g) / (159.69 g/mol) = 2.191 mol Fe₂O₃ According to the balanced equation, 1 mole of Fe₂O₃ reacts with 3 moles of CO, so we can find the moles of CO that can react with 2.191 moles of Fe₂O₃: moles of CO = 3 moles CO × 2.191 moles Fe₂O₃ = 6.573 mol CO To find the grams of CO: grams of CO = moles of CO × molar mass of CO = 6.573 mol × (28.01 g/mol) = 184.1 g CO

03

3. Calculate the grams of Fe and CO2 formed when 0.350 kg of Fe2O3 reacts

From the balanced equation, 1 mole of Fe₂O₃ produces 2 moles of Fe. So, 2.191 moles of Fe₂O₃ produce: moles of Fe = 2 moles Fe × 2.191 moles Fe₂O₃ = 4.382 mol Fe To find the grams of Fe: grams of Fe = moles of Fe × molar mass of Fe = 4.382 mol × 55.845 g/mol = 244.6 g Fe Now, calculate the moles of CO₂ formed: moles of CO₂ = 3 moles CO₂ × 2.191 moles Fe₂O₃ = 6.573 mol CO₂ To find the grams of CO₂: grams of CO₂ = moles of CO₂ × molar mass of CO₂ = 6.573 mol × 44.01 g/mol = 289.0 g CO₂

04

4. Verify if our calculations are consistent with the law of conservation of mass

The law of conservation of mass states that the mass of the reactants should be equal to the mass of the products in a chemical reaction. Mass of reactants = Mass of Fe₂O₃ + Mass of CO = 350 g + 184.1 g = 534.1 g Mass of products = Mass of Fe + Mass of CO₂ = 244.6 g + 289.0 g = 533.6 g The difference between the reactants and products is 0.5 g, which could be a rounding error. Therefore, our calculations are consistent with the law of conservation of mass.

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