Hi, Donald!
To solve this problem, we'll need to find the balance equation for this reaction. We already know the reactants and products from the problem:
Al(NO3)3 + NaCl --> NaNO3 + AlCl3
We need to balance this. I see that there are 3 NO3s on the reactants side, but only 1 on the products side. I can equalize this by adding a coefficient of 3 in front of NaNO3.
Al(NO3)3 + NaCl --> 3NaNO3 + AlCl3
Now, though, there are 3 Na on the products side and only 1 on the reactants side. I can equalize this by adding a coefficient of 3 to NaCl.
Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
Because I now have the same amount of each element or compound in both the reactants and products, my equation is balanced.
Next, we have to determine which of the reactants is the limiting reactant--the one that will run out first. It takes 3 times as much NaCl as it does Al(NO3)3 to complete this reaction, which we know from the coefficients. Therefore, the moles of NaCl used should be 3 times as many as the moles of Al(NO3)3 used. We know that we have 9 moles of NaCl. For these 9 moles to be used up completely, they'd have to react with 1/3 as much Al(NO3)3, or 3 moles. Because we have extra moles leftover of Al(NO3)3 (4 - 3 = 1), that means that it will not run out before NaCl, so NaCl is the limiting reactant.
Now that we know the limiting reactant, we can use dimensional analysis to go from moles of NaCl to moles of AlCl3 by using the coefficients in the reaction as a comparison point. I'll always put the units I want to move toward in the numerator of my comparisons and the units I want to move away from in the denominator.
9 mole NaCl (1 mole AlCl3 / 3 mole NaCl) = 3 mole AlCl3
Therefore, the maximum amount of AlCl3 that will result from this reaction is 3 moles.
I hope this helped!
