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?
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
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?