Answer:
9 moles of NaNO3.
Explanation:
The balanced equation for the reaction is given below:
Al(NO3)3 + 3NaCl —> 3NaNO3 + AlCl3
Next, we shall determine the number of mole of Al(NO3)3 and NaCl that reacted and the number of mole of NaNO3 produced from the balanced equation.
From the balanced equation above,
1 mole of Al(NO3)3 reacted with 3 moles of NaCl to produce 3 moles of NaNO3.
Next, we shall determine the limiting reactant.
The limiting reactant can be obtained as follow:
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 x 3)/1 = 12 moles of NaCl.
From the calculations made above, we can see that it will take a higher amount i.e 12 moles than what was given i.e 9 moles of NaCl to react completely with 4 moles of Al(NO3)3.
Therefore, NaCl is the limiting reactant and Al(NO3)3 is the excess reactant.
Finally, we shall determine the maximum amount of NaNO3 produced from the reaction.
In this case, the limiting reactant will be used as it will produce the maximum amount of NaNO3 since all of it is consumed by the reaction.
The limit reactant is NaCl and the maximum amount of NaNO3 produced can be obtained as follow:
From the balanced equation above,
3 moles of NaCl reacted to produce 3 moles of NaNO3.
Therefore, 9 moles of NaCl will also react to produce 9 moles of NaNO3.
From the calculations made above, the maximum amount of NaNO3 produced is 9 moles
![\frac{[H^+][A^-]}{[HA]} = 1.5 x 10^{-5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%3D%201.5%20x%2010%5E%7B-5%7D%20%20)
= 1.5 x 10⁻⁵