To determine how many grams of [tex]\(\text{Al(NO}_3\text{)}_3\)[/tex] can form during the reaction, follow these steps:
1. Understand the given information:
- Volume of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] solution: [tex]\(200.0 \, \text{mL}\)[/tex]
- Molarity of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] solution: [tex]\(0.500 \, \text{M}\)[/tex]
- Molar mass of [tex]\(\text{Al(NO}_3\text{)}_3\)[/tex]: [tex]\(213.01 \, \text{g/mol}\)[/tex]
- We previously determined that [tex]\(0.100 \, \text{mol Al}_2(\text{SO}_4)_3\)[/tex] reacts.
2. Determine the moles of [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] reacting:
- Given that we already know [tex]\(0.100 \, \text{mol Al}_2(\text{SO}_4)_3\)[/tex] is reacting.
3. Write and balance the chemical equation:
[tex]\[
3 \, \text{Ba(NO}_3\text{)}_2\text{(aq)} + \text{Al}_2(\text{SO}_4)_3\text{(aq)} \rightarrow 3 \, \text{BaSO}_4\text{(s)} + 2 \, \text{Al(NO}_3\text{)}_3\text{(aq)}
\][/tex]
4. Determine the mole ratio between [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex] and [tex]\(\text{Al(NO}_3\text{)}_3\)[/tex]:
- From the balanced equation, [tex]\(1 \, \text{mol Al}_2(\text{SO}_4)_3\)[/tex] produces [tex]\(2 \, \text{mol Al(NO}_3\text{)}_3\)[/tex].
5. Calculate the moles of [tex]\(\text{Al(NO}_3\text{)}_3\)[/tex] produced:
[tex]\[
\text{mol Al(NO}_3\text{)}_3 = 2 \times \text{mol Al}_2(\text{SO}_4)_3
\][/tex]
[tex]\[
\text{mol Al(NO}_3\text{)}_3 = 2 \times 0.100 \, \text{mol} = 0.200 \, \text{mol}
\][/tex]
6. Calculate the mass of [tex]\(\text{Al(NO}_3\text{)}_3\)[/tex] produced:
[tex]\[
\text{mass of Al(NO}_3\text{)}_3 = \text{mol Al(NO}_3\text{)}_3 \times \text{molar mass of Al(NO}_3\text{)}_3
\][/tex]
[tex]\[
\text{mass of Al(NO}_3\text{)}_3 = 0.200 \, \text{mol} \times 213.01 \, \text{g/mol}
\][/tex]
[tex]\[
\text{mass of Al(NO}_3\text{)}_3 \approx 42.60 \, \text{g}
\][/tex]
Therefore, the amount of [tex]\(\text{Al(NO}_3\text{)}_3\)[/tex] that can form during the reaction is approximately [tex]\(42.60 \, \text{g}\)[/tex].
