If 1.84 moles of Mg react, you can set up a proportion. 1.84/3 = x/2. solve for x = 1.23 moles aluminum produced
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
Dec 4th, 2014
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