3 moles of AlCl₃.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
Al(NO₃)₃ + 3NaCl —> 3NaNO₃ + AlCl₃
From the balanced equation above,
1 mole of Al(NO₃)₃ reacted with 3 moles of NaCl to 1 mole of AlCl₃.
Next, we shall determine the limiting reactant. This can be obtained as illustrated below:
From the balanced equation above,
1 mole of Al(NO₃)₃ reacted with 3 moles of NaCl.
Therefore, 4 moles of Al(NO₃)₃ will react with = 4 × 3 = 12 moles of NaCl.
From the calculations made above, a higher amount (i.e 12 moles) of NaCl than what was given (i.e 9 moles) is required to react completely with 4 moles of Al(NO₃)₃.
Thus, NaCl is the limiting reactant and Al(NO₃)₃. is the excess reactant.
Finally, we shall determine the maximum amount of AlCl₃ produced from the reaction.
The limiting reactant will be used to obtain the maximum amount of AlCl₃ because all of it is consumed by the reaction.
The limiting reactant is NaCl and the maximum amount of AlCl₃ produced can be obtained as follow:
From the balanced equation above,
3 moles of NaCl reacted to produce 1 mole of AlCl₃.
Therefore, 9 moles of NaCl will react to produce = (9 × 1)/3 = 3 moles of AlCl₃.
Thus, a maximum amount of 3 moles of AlCl₃ were obtained from the reaction.