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?
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
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?