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 chemical equation, we must ensure that there are the same number of atoms of each element on both sides of the equation. The initial equation is: $$ \mathrm{Fe}_{2}\mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ To balance the equation, we multiply the substances as needed: $$ \mathrm{Fe}_{2}\mathrm{O}_{3}(s)+3\mathrm{CO}(g) \longrightarrow 2\mathrm{Fe}(s)+3\mathrm{CO}_{2}(g) $$ Now, we have the same number of each type of atom on both sides of the equation.

We are given 0.350 kg of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) and need to find the mass of CO that can react with it. First, convert the mass of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) to moles using its molar mass (159.69 g/mol): $$ \text{moles of}\ \mathrm{Fe}_{2}\mathrm{O}_{3} = \frac{0.350\,\text{kg}\, \times\, 1000\,\text{g/kg}}{159.69\,\text{g/mol}} = 2.191\,\text{mol} $$ Now, use the stoichiometry of the balanced equation to find the moles of CO that react with 2.191 moles of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\): $$ \text{moles of CO} = 2.191\,\text{mol}\ \mathrm{Fe}_{2}\mathrm{O}_{3} \times \frac{3\,\text{mol}\,\mathrm{CO}}{1\,\text{mol}\,\mathrm{Fe}_{2}\mathrm{O}_{3}} = 6.573\,\text{mol}\,\mathrm{CO} $$ Finally, convert the moles of CO to grams using its molar mass (28.01 g/mol): $$ \text{mass of CO} = 6.573\,\text{mol}\,\mathrm{CO} \times 28.01\,\text{g/mol} = 184\,\text{g} $$ So, 184 grams of CO can react with 0.350 kg of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\).

To find the mass of Fe and CO\(_2\) formed when 0.350 kg of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) reacts, first find the moles of Fe produced using the stoichiometry of the balanced equation: $$ \text{moles of Fe} = 2.191\,\text{mol}\ \mathrm{Fe}_{2}\mathrm{O}_{3} \times \frac{2\,\text{mol}\,\mathrm{Fe}}{1\,\text{mol}\,\mathrm{Fe}_{2}\mathrm{O}_{3}} = 4.382\,\text{mol}\,\mathrm{Fe} $$ Now, convert the moles of Fe to grams using its molar mass (55.845 g/mol): $$ \text{mass of Fe} = 4.382\,\text{mol}\,\mathrm{Fe} \times 55.845\,\text{g/mol} = 245\,\text{g} $$ Next, find the moles of CO\(_2\) produced using the stoichiometry of the balanced equation: $$ \text{moles of CO}_{2} = 2.191\,\text{mol}\ \mathrm{Fe}_{2}\mathrm{O}_{3} \times \frac{3\,\text{mol}\,\mathrm{CO}_{2}}{1\,\text{mol}\,\mathrm{Fe}_{2}\mathrm{O}_{3}} = 6.573\,\text{mol}\,\mathrm{CO}_{2} $$ Then, convert the moles of CO\(_2\) to grams using its molar mass (44.01 g/mol): $$ \text{mass of CO}_{2} = 6.573\,\text{mol}\,\mathrm{CO}_{2} \times 44.01\,\text{g/mol} = 289\,\text{g} $$ So, 245 grams of Fe and 289 grams of CO\(_2\) are formed when 0.350 kg of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) reacts.

According to the law of conservation of mass, the total mass of reactants should be equal to the total mass of products. We will now verify this by comparing the mass of reactants with the mass of products: Mass of reactants = mass of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) + mass of CO Mass of reactants = 0.350 kg (350 g) + 184 g = 534 g Mass of products = mass of Fe + mass of CO\(_2\) Mass of products = 245 g + 289 g = 534 g Since the mass of reactants (534 g) is equal to the mass of products (534 g), our calculations are consistent with the law of conservation of mass.