We wish to determine how many grams of
aluminum nitrate can form when 200.0 mL
of 0.500 M aluminum sulfate reacts with excess
barium nitrate.
3Ba(NO3)2(aq) + Al₂(SO4)3(aq) → 3BaSO4(s) + 2AI(NO3)3(aq)
In the previous step, you determined
0.100 mol Al₂(SO4)3 react. The molar mass of
AI(NO3)3 is 213.01 g/mol.
How many grams of aluminum nitrate can form
during the reaction?
W
Mass (g) AI(NO)
Enter