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2Al(NO3)3 + 3Mg ==> 3Mg(NO3)2 + 2Al 

If 1.84 moles of magnesium are reacted, how many moles of aluminum will be produced?

What mass of aluminum nitrate is needed to produce 0.439 moles of magnesium nitrate?

Oct 23rd, 2017

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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Oct 23rd, 2017
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Oct 23rd, 2017
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