Questions: Calculate the number of grams of the product Fe2O3 that form. Be sure your answer has the correct number of significant figures. g Fe2O3 × 10

Solution Steps

To calculate the number of grams of \(\mathrm{Fe}_2\mathrm{O}_3\) formed, we need the balanced chemical equation for the reaction. Assuming the reaction is the oxidation of iron:

\[ 4 \mathrm{Fe} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{Fe}_2\mathrm{O}_3 \]

The molar mass of \(\mathrm{Fe}_2\mathrm{O}_3\) can be calculated as follows:

• Molar mass of Fe: \(55.845 \, \text{g/mol}\)
• Molar mass of O: \(15.999 \, \text{g/mol}\)

\[ \text{Molar mass of } \mathrm{Fe}_2\mathrm{O}_3 = 2 \times 55.845 + 3 \times 15.999 = 111.69 + 47.997 = 159.687 \, \text{g/mol} \]

Assuming we start with a certain amount of iron (Fe), we need to convert this to moles, use the stoichiometric ratio from the balanced equation, and then convert back to grams of \(\mathrm{Fe}_2\mathrm{O}_3\).

For example, if we start with \(x\) grams of Fe:

1. Convert grams of Fe to moles: \[ \text{Moles of Fe} = \frac{x \, \text{g}}{55.845 \, \text{g/mol}} \]

2. Use the stoichiometric ratio to find moles of \(\mathrm{Fe}_2\mathrm{O}_3\): \[ \text{Moles of } \mathrm{Fe}_2\mathrm{O}_3 = \frac{1}{4} \times \text{Moles of Fe} = \frac{1}{4} \times \frac{x}{55.845} \]

3. Convert moles of \(\mathrm{Fe}_2\mathrm{O}_3\) to grams: \[ \text{Grams of } \mathrm{Fe}_2\mathrm{O}_3 = \text{Moles of } \mathrm{Fe}_2\mathrm{O}_3 \times 159.687 \, \text{g/mol} = \left( \frac{1}{4} \times \frac{x}{55.845} \right) \times 159.687 \]

Final Answer