Answer:
The answer to your question is 3 moles of AlCl₃
Explanation:
Process
1.- Write and balance the equation
Al(NO₃)₃ + 3NaCl ⇒ 3NaNO₃ + AlCl₃
2.- Determine the limiting reactant
Theoretical proportion 1 mol Al(NO₃)₃ : 3 moles of NaCl
Experimental proportion 4 moles Al(NO₃)₃ : 9 moles NaCl
From these values, we determine that the limiting reactant is NaCl because the number of moles increases three times and the number of moles of Al(NO₃)₃ increases four times.
3.- Determine the amount of AlCl₃ using proportions
3 moles of NaCl --------------- 1 mol of AlCl₃
9 moles of NaCl ---------------- x
x = (9 x 1) / 3
x = 9 /3
x = 3 moles
