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…

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?

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?

Answer

Explanation:

Step1: Determine moles of ( \text{Al(NO}_3\text{)}_3 )

From the balanced equation, 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 )

Use the formula ( \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 g (or 42.602 g, depending on precision)