Question

we wish to determine 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. 3ba(no₃)₂(aq) + al₂(so₄)₃(aq) → 3baso₄(s) + 2al(no₃)₃(aq) in the previous step, you determined 0.100 mol al₂(so₄)₃ react. the molar mass of al(no₃)₃ is 213.01 g/mol. how many grams of al(no₃)₃ can form during the reaction?

Explanation:

Step1: Find moles of \( \text{Al(NO}_3\text{)}_3 \) from stoichiometry

From the balanced equation \( 3\text{Ba(NO}_3\text{)}_2 + \text{Al}_2(\text{SO}_4)_3
ightarrow 3\text{BaSO}_4 + 2\text{Al(NO}_3\text{)}_3 \), 1 mol of \( \text{Al}_2(\text{SO}_4)_3 \) produces 2 mol of \( \text{Al(NO}_3\text{)}_3 \). Given 0.100 mol of \( \text{Al}_2(\text{SO}_4)_3 \) reacts, moles of \( \text{Al(NO}_3\text{)}_3 = 0.100 \, \text{mol} \times 2 = 0.200 \, \text{mol} \).

Step2: Calculate mass of \( \text{Al(NO}_3\text{)}_3 \)

Using \( \text{mass} = \text{moles} \times \text{molar mass} \), molar mass of \( \text{Al(NO}_3\text{)}_3 \) is 213.01 g/mol. So mass \( = 0.200 \, \text{mol} \times 213.01 \, \text{g/mol} = 42.602 \, \text{g} \).

Answer:

\( 42.6 \, \text{g} \) (or 42.602 g, depending on precision)