Answer :
To determine the excess reactant in the reaction given by the balanced equation:
[tex]$ 3 \, \text{Ba} + \text{Al}_2(\text{SO}_4)_3 \rightarrow 2 \, \text{Al} + 3 \, \text{BaSO}_4 $[/tex]
where we have 0.65 moles of [tex]\(\text{Ba}\)[/tex] and 0.50 moles of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex], we need to follow these steps:
1. Identify the mole ratio from the balanced equation:
- According to the balanced equation, 3 moles of [tex]\(\text{Ba}\)[/tex] react with 1 mole of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex].
2. Calculate the moles of [tex]\(\text{Ba}\)[/tex] required to completely react with 0.50 moles of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex]:
- For 1 mole of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex], 3 moles of [tex]\(\text{Ba}\)[/tex] are required.
- Therefore, for 0.50 moles of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex]:
[tex]$ \text{Moles of Ba required} = 0.50 \, \text{moles of } \text{Al}_2(\text{SO}_4)_3 \times 3 = 1.5 \, \text{moles of Ba} $[/tex]
3. Compare the moles of [tex]\(\text{Ba}\)[/tex] available to the moles of [tex]\(\text{Ba}\)[/tex] required:
- We have 0.65 moles of [tex]\(\text{Ba}\)[/tex] available.
- We need 1.5 moles of [tex]\(\text{Ba}\)[/tex] to fully react with 0.50 moles of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex].
4. Determine the limiting reactant and the excess reactant:
- Since we only have 0.65 moles of [tex]\(\text{Ba}\)[/tex], which is less than the 1.5 moles required, [tex]\(\text{Ba}\)[/tex] is the limiting reactant.
- This means [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] will be the excess reactant.
5. Calculate the amount of the excess reactant remaining after the reaction:
- First, calculate how much [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] will react with the available [tex]\(\text{Ba}\)[/tex]:
[tex]$ \text{Moles of } \text{Al}_2(\text{SO}_4)_3 \text{ reacting} = \frac{0.65 \, \text{moles of Ba}}{3} = \frac{0.65}{3} \approx 0.2167 \text{ moles of } \text{Al}_2(\text{SO}_4)_3 $[/tex]
- Subtract the moles of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] that have reacted from the initial amount:
[tex]$ \text{Excess amount of } \text{Al}_2(\text{SO}_4)_3 = 0.50 \, \text{moles} - 0.2167 \, \text{moles} = 0.2833 \, \text{moles} $[/tex]
Therefore, the excess reactant is [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex], and the amount of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] that remains unreacted is approximately 0.2833 moles.
[tex]$ 3 \, \text{Ba} + \text{Al}_2(\text{SO}_4)_3 \rightarrow 2 \, \text{Al} + 3 \, \text{BaSO}_4 $[/tex]
where we have 0.65 moles of [tex]\(\text{Ba}\)[/tex] and 0.50 moles of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex], we need to follow these steps:
1. Identify the mole ratio from the balanced equation:
- According to the balanced equation, 3 moles of [tex]\(\text{Ba}\)[/tex] react with 1 mole of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex].
2. Calculate the moles of [tex]\(\text{Ba}\)[/tex] required to completely react with 0.50 moles of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex]:
- For 1 mole of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex], 3 moles of [tex]\(\text{Ba}\)[/tex] are required.
- Therefore, for 0.50 moles of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex]:
[tex]$ \text{Moles of Ba required} = 0.50 \, \text{moles of } \text{Al}_2(\text{SO}_4)_3 \times 3 = 1.5 \, \text{moles of Ba} $[/tex]
3. Compare the moles of [tex]\(\text{Ba}\)[/tex] available to the moles of [tex]\(\text{Ba}\)[/tex] required:
- We have 0.65 moles of [tex]\(\text{Ba}\)[/tex] available.
- We need 1.5 moles of [tex]\(\text{Ba}\)[/tex] to fully react with 0.50 moles of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex].
4. Determine the limiting reactant and the excess reactant:
- Since we only have 0.65 moles of [tex]\(\text{Ba}\)[/tex], which is less than the 1.5 moles required, [tex]\(\text{Ba}\)[/tex] is the limiting reactant.
- This means [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] will be the excess reactant.
5. Calculate the amount of the excess reactant remaining after the reaction:
- First, calculate how much [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] will react with the available [tex]\(\text{Ba}\)[/tex]:
[tex]$ \text{Moles of } \text{Al}_2(\text{SO}_4)_3 \text{ reacting} = \frac{0.65 \, \text{moles of Ba}}{3} = \frac{0.65}{3} \approx 0.2167 \text{ moles of } \text{Al}_2(\text{SO}_4)_3 $[/tex]
- Subtract the moles of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] that have reacted from the initial amount:
[tex]$ \text{Excess amount of } \text{Al}_2(\text{SO}_4)_3 = 0.50 \, \text{moles} - 0.2167 \, \text{moles} = 0.2833 \, \text{moles} $[/tex]
Therefore, the excess reactant is [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex], and the amount of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] that remains unreacted is approximately 0.2833 moles.
