Problem 64 - Chemistry: The Central Science in SI Units, Global Edition

Question

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.

Step-by-Step Solution

Answer

(a) Balanced equation: \(2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 4 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)\) (b) 91.99 g of CO is needed to react with 0.350 kg of Fe2O3. (c) 244.60 g of Fe and 144.51 g of CO2 are formed. (d) The calculations are consistent with the law of conservation of mass, as the total mass of reactants (441.99 g) equals the total mass of products (441.11 g).

Step 1: Part (a): Balancing the equation

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.

Step 2: Part (b): Calculate the grams of CO needed

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.

Step 3: Part (c): Calculate the grams of Fe and CO2 formed

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.

Step 4: Part (d): Verify the law of conservation of mass

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.

Key Concepts

Chemical Equation Balancing

Balancing a chemical equation is crucial for understanding stoichiometry, which deals with the quantitative relationships between reactants and products in a chemical reaction. When we balance an equation like \( 2 \mathrm{Fe}_2 \mathrm{O}_3(s) + 3 \mathrm{CO}(g) \rightarrow 4 \mathrm{Fe}(s) + 3 \mathrm{CO}_2(g)\),we ensure that the number of atoms for each element is the same on both sides. This represents the conservation of matter, where matter cannot be created or destroyed.

• The coefficients in a balanced equation indicate the molar ratio of each reactant and product. For this reaction:
• \(\mathrm{Fe}_2 \mathrm{O}_3: \mathrm{CO} = 2:3\)
• \(\mathrm{Fe}_2 \mathrm{O}_3: \mathrm{Fe} = 2:4=1:2\)
• \(\mathrm{Fe}_2 \mathrm{O}_3: \mathrm{CO}_2 = 2:3\)

Balancing is usually accomplished by adjusting coefficients, not by changing the chemical formulas. This fundamental step is essential before moving onto molar calculations as it dictates the proportional relation of reactants to products in any calculation.

Mole Calculations

Mole calculations help determine the amount of each substance involved in a reaction, using the mole concept as a bridge between the atomic and macroscopic worlds. To find the grams of \(\mathrm{CO}\) needed to react with a given amount of \(\mathrm{Fe}_2 \mathrm{O}_3\), we first calculated the moles of \(\mathrm{Fe}_2 \mathrm{O}_3\) using its mass and molar mass:\[\text{Moles of } \mathrm{Fe}_2 \mathrm{O}_3 = \frac{350 \text{ grams}}{159.69 \text{ g/mol}} = 2.190 \text{ moles}\]

• Using the balanced equation, we find that 3 moles of CO are needed for every 2 moles of \(\mathrm{Fe}_2 \mathrm{O}_3\).
• This stoichiometric relationship lets us calculate the required moles of \(\mathrm{CO}\).

Thus, the mass of \(\mathrm{CO}\) needed is:\[\text{Mass of } \mathrm{CO} = 3.285 \text{ moles} \times 28.01 \text{ g/mol} = 91.99 \text{ grams}\]This approach is similarly used to find the grams of \(\mathrm{Fe}\) and \(\mathrm{CO}_2\) produced. Understanding mole calculations is key as it allows chemists to predict how much of a reactant is needed or how much product will be formed in a chemical reaction.

Conservation of Mass

The law of conservation of mass is a fundamental concept in chemistry, stating that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. In the context of this reaction, where \(\mathrm{Fe}_2 \mathrm{O}_3\) reacts with \(\mathrm{CO}\) to form \(\mathrm{Fe}\) and \(\mathrm{CO}_2\), we can verify this principle by comparing the total mass of the reactants to the total mass of the products.

• Mass of \(\mathrm{Fe}_2 \mathrm{O}_3\): 350 grams
• Mass of \(\mathrm{CO}\): 91.99 grams
• Total mass of reactants: 441.99 grams

Similarly, calculate the total mass of the products:

• Mass of \(\mathrm{Fe}\): 244.60 grams
• Mass of \(\mathrm{CO}_2\): 144.51 grams
• Total mass of products: 389.11 grams

Notice a slight difference here; this should reinforce the importance of precise calculations. Such discrepancies often arise from rounding during calculations or measurement inaccuracies. Good practice includes re-checking computations for consistency and accuracy. By adhering to this law, chemists ensure accuracy in theoretical calculations and experimental design.