To calculate the amount of heat transferred when [tex]\(5.5 \, \text{g}\)[/tex] of aluminum ([tex]\( \text{Al} \)[/tex]) is consumed during this reaction, we can follow a few steps.
### Step 1: Determine the number of moles of Al
First, let's convert the mass of aluminum into moles. We use the molar mass of aluminum, which is [tex]\(26.98 \, \text{g/mol}\)[/tex].
[tex]\[ \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} \][/tex]
Given that the mass of Al is [tex]\(5.5 \, \text{g}\)[/tex]:
[tex]\[ \text{moles of Al} = \frac{5.5 \, \text{g}}{26.98 \, \text{g/mol}} \approx 0.20385 \, \text{mol} \][/tex]
### Step 2: Calculate the heat transfer per mole of Al
According to the balanced chemical equation, 4 moles of Al produce [tex]\(-1669.8 \, \text{kJ}\)[/tex] of heat. So, we need to find the amount of heat produced per mole of Al.
[tex]\[ \text{heat per mole of Al} = \frac{-1669.8 \, \text{kJ}}{4 \, \text{mol}} = -417.45 \, \text{kJ/mol} \][/tex]
### Step 3: Calculate the total heat transferred for 5.5 g of Al
Now, we need to multiply the moles of Al by the heat produced per mole to find out the total heat transferred for the 5.5 g of Al.
[tex]\[ \text{heat transferred} = \text{moles of Al} \times \text{heat per mole of Al} \][/tex]
[tex]\[ \text{heat transferred} = 0.20385 \, \text{mol} \times -417.45 \, \text{kJ/mol} = -85.10 \, \text{kJ} \][/tex]
### Conclusion
The amount of heat transferred when [tex]\(5.5 \, \text{g}\)[/tex] of aluminum is consumed during this reaction is approximately [tex]\(-85.10 \, \text{kJ}\)[/tex].
