1. **State the problem:** We want to find how many grams of Al(NO3)3 can form when 200.0 mL of 0.500 M Al2(SO4)3 reacts with excess Ba(NO3)2. 2. **Given reaction:** $$3\text{Ba(NO}_3)_2 + \text{Al}_2(\text{SO}_4)_3 \rightarrow 3\text{BaSO}_4 + 2\text{Al(NO}_3)_3$$ 3. **Step 1: Calculate moles of Al2(SO4)3 present.** Molarity (M) = moles/volume(L), so $$\text{moles} = M \times V = 0.500 \times 0.200 = 0.100\text{ moles}$$ 4. **Step 2: Use stoichiometry to find moles of Al(NO3)3 formed.** From the balanced equation, 1 mole of Al2(SO4)3 produces 2 moles of Al(NO3)3. So, $$\text{moles Al(NO}_3)_3 = 0.100 \times 2 = 0.200$$ 5. **Step 3: Calculate mass of Al(NO3)3 formed.** Molar mass of Al(NO3)3: - Al: 26.98 g/mol - N: 14.01 g/mol - O: 16.00 g/mol Calculate: $$26.98 + 3 \times (14.01 + 3 \times 16.00) = 26.98 + 3 \times (14.01 + 48.00) = 26.98 + 3 \times 62.01 = 26.98 + 186.03 = 213.01\text{ g/mol}$$ 6. **Step 4: Calculate mass:** $$\text{mass} = \text{moles} \times \text{molar mass} = 0.200 \times 213.01 = 42.60\text{ g}$$ **Final answer:** $$\boxed{42.60\text{ grams of Al(NO}_3)_3}$$