Questions: How much heat is produced when 68.5 grams of Fe2O3 react? Be sure your answer has the correct number of significant figures.

Solution Steps

First, we need to calculate the molar mass of \(\mathrm{Fe}_2\mathrm{O}_3\). The molar masses of iron (Fe) and oxygen (O) are approximately 55.845 g/mol and 16.00 g/mol, respectively.

\[ \text{Molar mass of } \mathrm{Fe}_2\mathrm{O}_3 = 2 \times 55.845 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol} = 111.69 \, \text{g/mol} + 48.00 \, \text{g/mol} = 159.69 \, \text{g/mol} \]

Next, we calculate the number of moles of \(\mathrm{Fe}_2\mathrm{O}_3\) in 68.5 grams.

\[ \text{Number of moles} = \frac{68.5 \, \text{g}}{159.69 \, \text{g/mol}} = 0.4291 \, \text{mol} \]

From the problem, we know that the reaction produces \(1.06 \times 10^3 \, \text{kJ}\) per mole of \(\mathrm{Fe}_2\mathrm{O}_3\).

Finally, we calculate the total heat produced when 68.5 grams of \(\mathrm{Fe}_2\mathrm{O}_3\) react.

\[ \text{Total heat produced} = 0.4291 \, \text{mol} \times 1.06 \times 10^3 \, \text{kJ/mol} = 454.846 \, \text{kJ} \]

Since the number of significant figures in the given mass (68.5 grams) is three, we should round our final answer to three significant figures.

\[ \text{Total heat produced} \approx 455 \, \text{kJ} \]

Final Answer