Reaction Data Reactants Products

Al(NO3)3 NaCl NaNO3 AlCl3

Starting Amount in Reaction 4 moles 9 moles ? ?

Determine the maximum amount of NaNO3 that was produced during the experiment. Explain how you determined this amount. (5 points)

9 moles of NaNO₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated 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 produce 3 moles of NaNO₃.

Next, we shall determine the limiting reactant. This 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 × 3 = 12 moles of NaCl.

From the calculation made above, we can see that it will take a higher amount (i.e 12 moles) of NaCl than what was given (i.e 9 moles) 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 use because it will produce the maximum amount of NaNO₃ since all of it is used up in the reaction.

The limiting reactant is NaCl and 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

Thus, 9 moles of NaNO₃ were obtained from the reaction.