The equation is already stoichiometrically balanced, making this problem more simple.

The molar mass of aluminum is 26.982 g, so 54.1/26.982 = 2.005 moles Al produced, which we can round to 2 moles to further simplify the problem. Since 3 moles of aluminum nitrate produces 2 moles of aluminum, we can conclude that 3 moles of aluminum nitrate is reacted.

The next part of the problem involves a proportion. For every 2 moles of aluminum nitrate reacted, 3 moles of magnesium nitrate are produced. So you can set up the proportion as: x/2 = 0.439/3. Solve for x: x = 0.293 moles aluminum nitrate. Multiply this by the molar mass of aluminum nitrate ((2 x 29.982) + (3 x 14) + (9 x 16) = 240.1). So 0.293 x 240.1 = 70.35 g aluminum nitrate is needed