Answer:
9 moles of NaNO3 and 3 moles of AlCl3
Explanation:
First, we need the balanced equation, which is:
Al(NO3)3 + (3)NaCl ---> (3)NaNO3 + AlCl3
Next, we need the limiting reactant, which is NaCl because we require more of it than Al(NO3) to make AICl
Now we need to determine the maximum of AlCl we can produce:
Because we need 3 moles of NaCl to react with Al(NO3)3 to produce 1 mole of AlCl3, we can only produce 3 AlCl3 due to insufficient NaCl.
So we would have 9 NaNO3 and 3 AlCl3.