Question\text{How much heat is produced when 74 grams of Al react in the following reaction?} \\ 2 \mathrm{Al}(s) + \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \rightarrow 2 \mathrm{Fe}(s) + \mathrm{Al}_{2} \mathrm{O}_{3}(s) \\ \text{Heat of reaction} = -851.5 \, \mathrm{kJ} \\

Studdy Solution

STEP 1

1. The balanced chemical equation is $2 \text{Al}(s) + \text{Fe}_2\text{O}_3(s) \rightarrow 2 \text{Fe}(s) + \text{Al}_2\text{O}_3(s)$ .
2. The heat of reaction is $-851.5 \, \text{kJ}$ .
3. We are given 74 grams of aluminum ( $\text{Al}$ ).
4. We need to calculate the heat produced when 74 grams of $\text{Al}$ react.

STEP 2

1. Calculate the molar mass of aluminum.
2. Determine the number of moles of aluminum in 74 grams.
3. Use stoichiometry to find the heat produced.

STEP 3

Calculate the molar mass of aluminum ( $\text{Al}$ ).
The atomic mass of aluminum is approximately $26.98 \, \text{g/mol}$ .

STEP 4

Determine the number of moles of aluminum in 74 grams.
Use the formula: $\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}}$
Substitute the known values: $\text{Number of moles of Al} = \frac{74 \, \text{g}}{26.98 \, \text{g/mol}} \approx 2.74 \, \text{mol}$

STEP 5

Use stoichiometry to find the heat produced.
According to the balanced equation, 2 moles of $\text{Al}$ produce $-851.5 \, \text{kJ}$ of heat.
Calculate the heat produced for 2.74 moles of $\text{Al}$ : $\text{Heat produced} = \left(\frac{-851.5 \, \text{kJ}}{2 \, \text{mol}}\right) \times 2.74 \, \text{mol}$ $\text{Heat produced} \approx -1166.5 \, \text{kJ}$
The heat produced when 74 grams of aluminum react is approximately $-1166.5 \, \text{kJ}$ .

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