Chemistry question

Question

How many g of Al(NO3)3 do we get after the reaction of 10,2 g Al2O3 with 200 ml HNO3 solution (3 mol/l)?

Answer 1

Start with a correctly balanced equation:

Al2O3 + 6HNO3 ==> 2Al(NO3)3 + 3H2O

Next, find the limiting reactant:

moles Al2O3 = 10.2 g Al2O3 x 1 mol Al2O3/101.96 g = 0.100 moles Al2O3 present

moles HNO3 = 0.2 L x 3 mol/L = 0.6 mol HNO3 present

Since both reactants are present in their stoichiometric amounts, neither is limiting. Using Al2O3, we get....

0.1 moles Al2O3 x 2 mol Al(NO3)3 x 213 g/mol = 42.6 g Al(NO3)3 produced

Answer 2