can someone help please

Chemistry
Tutor: None Selected Time limit: 1 Day

2Al(NO3)3 + 3Mg ==> 3Mg(NO3)2 + 2Al 

A, If 21.2 g of aluminum is produced, how many moles of aluminum nitrate is reacted?

 B.What mass of aluminum nitrate is needed to produce 0.625 moles of magnesium nitrate? 

Dec 4th, 2014

A. The molar mass of Al is 27 g/mol.

     So 21.2g Al=21.2/27=0.785 mol Al

1 mol of Al(NO3)3 will produce 1 mol of Al, so when 0.785 mol of Al is produced, 0.785 mol of Al(NO3)3 is reacted.

B. The molar ration between Al(NO3)3 and Mg(NO3)2 is 2:3.

     So to produce 0.625 moles of magnesium nitrate, 0.625x2/3=0.417 mol of aluminum nitrate is needed.

    Therefore the mass of aluminum nitrate needed is 0.417x213=88.7 g

Dec 4th, 2014

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