To find the number of moles of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex] present in 50.0 mL of [tex]\( 0.250 \)[/tex] M [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex], follow these steps:
1. Convert the volume from mL to L:
1 L = 1000 mL, so:
[tex]\[
50.0 \, \text{mL} = 50.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.0500 \, \text{L}
\][/tex]
2. Use the molarity formula to find the moles of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex]:
Molarity (M) is defined as moles of solute per liter of solution. Therefore, the number of moles of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex] can be calculated using:
[tex]\[
\text{Moles of } \text{Al}_2(\text{SO}_4)_3 = \text{Molarity} \times \text{Volume (in liters)}
\][/tex]
Substituting the given values:
[tex]\[
\text{Moles of } \text{Al}_2(\text{SO}_4)_3 = 0.250 \, \text{M} \times 0.0500 \, \text{L} = 0.0125 \, \text{moles}
\][/tex]
So, there are [tex]\( 0.0125 \)[/tex] moles of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex] present in 50.0 mL of a [tex]\( 0.250 \)[/tex] M solution of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex].
Next, to determine how many moles of [tex]\( \text{BaSO}_4 \)[/tex] are formed, we use the chemical reaction provided:
[tex]\[
3 \, \text{Ba(NO}_3\text{)}_2 (aq) + \text{Al}_2(\text{SO}_4)_3 (aq) \rightarrow 3 \, \text{BaSO}_4 (s) + 2 \, \text{Al(NO}_3\text{)}_3 (aq)
\][/tex]
According to the stoichiometry of the balanced chemical equation, 1 mole of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex] produces 3 moles of [tex]\( \text{BaSO}_4 \)[/tex].
Given we have [tex]\( 0.0125 \)[/tex] moles of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex]:
[tex]\[
\text{Moles of } \text{BaSO}_4 = 3 \times \text{Moles of } \text{Al}_2(\text{SO}_4)_3 = 3 \times 0.0125 = 0.0375 \, \text{moles}
\][/tex]
Therefore, [tex]\( 0.0375 \)[/tex] moles of [tex]\( \text{BaSO}_4 \)[/tex] are formed when 50.0 mL of [tex]\( 0.250 \)[/tex] M [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex] reacts with excess [tex]\( \text{Ba(NO}_3\text{)}_2 \)[/tex].
