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2Al(NO3)3 + 3Mg ==> 3Mg(NO3)2 + 2Al

If 54.1 g of aluminum is produced, how many moles of aluminum nitrate is reacted?

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

Dec 4th, 2014

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

Dec 4th, 2014

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Dec 4th, 2014
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Dec 4th, 2014
Aug 19th, 2017
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