We wish to determine how many grams of [tex]\[Al \left( NO _3\right)_3\][/tex] can form when 200.0 mL of [tex]\[0.500 M Al _2\left( SO _4\right)_3\][/tex] reacts with excess [tex]\[Ba \left( NO _3\right)_2\][/tex].

[tex]\[3 Ba \left( NO _3\right)_2( aq )+ Al _2\left( SO _4\right)_3( aq ) \rightarrow 3 BaSO _4(s)+2 Al \left( NO _3\right)_3( aq )\][/tex]

How many moles of [tex]\[Al _2\left( SO _4\right)_3\][/tex] are present in 200.0 mL of [tex]\[0.500 M Al _2\left( SO _4\right)_3\][/tex]?



Answer :

To determine how many moles of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex] are present in 200.0 mL of a 0.500 M solution of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex], we need to follow these steps:

1. Convert the volume from milliliters (mL) to liters (L):
[tex]\[ 200.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.200 \text{ L} \][/tex]

2. Use the definition of molarity (M) which is moles of solute per liter of solution:
[tex]\[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \][/tex]

3. Calculate the number of moles of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex]:
Given the molarity of the solution is 0.500 M, and the volume of the solution is 0.200 L, the number of moles of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex] is:
[tex]\[ \text{moles of } \text{Al}_2(\text{SO}_4)_3 = 0.500 \text{ M} \times 0.200 \text{ L} = 0.100 \text{ moles} \][/tex]

Therefore, there are 0.100 moles of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex] present in 200.0 mL of a 0.500 M solution.