91Ó°ÊÓ

To balance the equation, begin by observing that iron (Fe), carbon (C), and oxygen (O) should be balanced on both sides of the equation. The unbalanced equation is:\[\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\]Firstly, balance Fe atoms by putting a coefficient of 2 in front of \(\mathrm{Fe}\):\[\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2\mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\]Next, balance the oxygen by placing a coefficient of 3 in front of \(\mathrm{CO}_{2}\):\[\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2\mathrm{Fe}(s)+3\mathrm{CO}_{2}(g)\]Finally, balance the carbon by placing a coefficient of 3 in front of \(\mathrm{CO}\):\[\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3\mathrm{CO}(g) \longrightarrow 2\mathrm{Fe}(s)+3\mathrm{CO}_{2}(g)\]Now, the chemical equation is balanced.

Convert the mass of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) from kilograms to grams: 0.350 kg = 350 g. The molar mass of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) is calculated as follows:\[\text{Molar mass of } \mathrm{Fe}_{2}\mathrm{O}_{3} = 2(55.85) + 3(16.00) = 159.70 \text{ g/mol}\]Calculate the moles of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\):\[\text{Moles of } \mathrm{Fe}_{2}\mathrm{O}_{3} = \frac{350 \text{ g}}{159.70 \text{ g/mol}} \approx 2.19 \text{ moles}\]

From the balanced equation:\[\mathrm{Fe}_{2}\mathrm{O}_{3} + 3\mathrm{CO} \longrightarrow 2\mathrm{Fe} + 3\mathrm{CO}_{2}\]1 mole of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) reacts with 3 moles of \(\mathrm{CO}\). Therefore, 2.19 moles of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) react with:\[2.19 \times 3 = 6.57 \text{ moles of } \mathrm{CO}\]The molar mass of \(\mathrm{CO}\) is \(12.01 + 16.00 = 28.01 \text{ g/mol}\).Calculate the mass of \(\mathrm{CO}\) used:\[\text{Mass of } \mathrm{CO} = 6.57 \text{ moles} \times 28.01 \text{ g/mol} \approx 184.13 \text{ g}\]

2 moles of \(\mathrm{Fe}\) are formed from 1 mole of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\). Therefore, 2.19 moles of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) will form:\[2.19 \times 2 = 4.38 \text{ moles of } \mathrm{Fe}\]The molar mass of \(\mathrm{Fe}\) is 55.85 g/mol.Calculate the mass of \(\mathrm{Fe}\):\[\text{Mass of } \mathrm{Fe} = 4.38 \text{ moles} \times 55.85 \text{ g/mol} \approx 244.63 \text{ g}\]

From the reaction, 3 moles of \(\mathrm{CO}_{2}\) are produced for every 1 mole of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\). Therefore, 2.19 moles of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) will produce:\[2.19 \times 3 = 6.57 \text{ moles of } \mathrm{CO}_{2}\]The molar mass of \(\mathrm{CO}_{2}\) is \(12.01 + 2 \times 16.00 = 44.01 \text{ g/mol}\).Calculate the mass of \(\mathrm{CO}_{2}\):\[\text{Mass of } \mathrm{CO}_{2} = 6.57 \text{ moles} \times 44.01 \text{ g/mol} \approx 289.31 \text{ g}\]

The law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction. Calculate the total mass of the reactants and products to confirm this principle:- Total initial mass: 350 g \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) + 184.13 g \(\mathrm{CO}\) = 534.13 g.- Total final mass: 244.63 g \(\mathrm{Fe}\) + 289.31 g \(\mathrm{CO}_{2}\) = 533.94 g.The total initial and final masses are approximately equal, confirming the conservation of mass within approximation errors due to rounding.