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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?

If 39.9 g of aluminum nitrate is reacted, what mass of magnesium nitrate is produced?

Dec 4th, 2014

For the first part of the question, use the ratios of the moles of magnesium to aluminum.  You need to fix the stoichiometry of the equation so that the molecules in the products are in the same amount as in the reactants. The equation is already stoichiometrically balanced, so this has been done for you already.  You can see that for every 3 moles of Mg, 2 moles of Al are produced.  So set up a proportion.  1.84/3 = x/2.  Solve for x, and x = 1.23.  So 1.23 moles of aluminum will be produced for every 1.84 moles of magnesium.

For the second part of the question, you will need the molar masses of all of the elements (which you can find using a periodic table).  Al = 26.98, N = 14, O = 16, and Mg = 24.3.  So the molar mass of aluminum nitrate is (2 x 26.98) + (3 x 14) + (9 x 16) = 239.96.  (39.9/239.96) gives you the number of moles of aluminum nitrate, so that is equal to 0.166 moles.  Since there are 3 moles Mg(NO3)2 for every 2 moles of aluminum nitrate, you can set up a proportion: 0.166/2 = x/3.  x = 0.249 moles magnesium nitrate.  Find the molar mass of magnesium nitrate: (3 x 24.3) + (2 x 14) + (6 x 16) = 196.9.  Now multiply the molar mass by the number of moles: 196.9 x 0.249 = 49.03 g magnesium nitrate produced

Dec 4th, 2014

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Dec 4th, 2014
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Dec 4th, 2014
Oct 22nd, 2017
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