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2Al(NO3)3 + 3Mg ==> 3Mg(NO3)2 + 2Al 

A. If 3.54 moles of magnesium are reacted, how many moles of aluminum will be produced? 

B. If 97.6 g of aluminum nitrate is reacted, what mass of magnesium nitrate is produced? 

Oct 18th, 2017

The molar ration between Al and Mg is 2:3.

If 3.54 moles of magnesium are reacted, 3.54x2/3=2.36 moles of aluminum will be produced.

The molar mass of Al(NO3)3=27+(14+16x3)x3=213 g/mol

So 97.6g Al(NO3)3=97.6/213=0.458 mol Al(NO3)3

The molar ratio of Mg(NO3)2 and Al(NO3)3 is 3:2.

If 0.458 mol Al(NO3)3 is reacted, 0.458x3/2=0.687 mol Mg(NO3)2 is produced.

The molar mass of Mg(NO3)2=24+(14+16x3)x2=148 g/mol

Therefore 0.687x148=101.7g Mg(NO3)2 is produced

Dec 4th, 2014

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