30cm³ of 0.1 M AI(NO3)3 solution is reacted with100cm³ of 0.15M of NaOH solution. Which is in excess and by how much?

A. NaOH solution, by 70cm³

B. NaOH solution, by 60cm³

C. NaOH solution by 40cm³

D. AI (NO3)3, solution by 20cm³

E. AI (NO3)3 solution, by 10cm³

Al(NO3)3 + 3NaOH ==> Al(OH)3 + 3NaNO3

mols Al(NO)3 = M x L = 0.1 x 0.03 = 0.003
mol NaOH = 0.15 x 0.1 = 0.015

It will take how much NaOH to react completely with the Al(NO3)3? That's 0.003 x 3 = 0.009. Do you have that much? yes. So Al(OH)3 is the limiting reagent and NaOH is the reagent in excess. How much excess? That's 0.015-0.009 = 0.006 mols NaOH too much.
M = mol/L = 0.15 = 0.006/L and solve for L and convert to cc.

Is it safe to say option C is correct?

To determine which reactant is in excess and by how much, we can use the concept of stoichiometry and the reaction equation.

The balanced chemical equation for the reaction between AI(NO3)3 and NaOH is:

AI(NO3)3 + 3NaOH → AI(OH)3 + 3NaNO3

From the equation, we can see that the ratio of AI(NO3)3 to NaOH is 1:3.

First, let's calculate the number of moles of AI(NO3)3 and NaOH in the given volumes.

Number of moles of AI(NO3)3 = Volume (in L) * Molarity
= 30 cm³ * (1 L / 1000 cm³) * 0.1 M
= 0.003 moles

Number of moles of NaOH = Volume (in L) * Molarity
= 100 cm³ * (1 L / 1000 cm³) * 0.15 M
= 0.015 moles

Now, let's compare the mole ratios of AI(NO3)3 to NaOH.

The mole ratio of AI(NO3)3 to NaOH is 1:3.

Since the mole ratio is 1:3, it means that for every 1 mole of AI(NO3)3, we need 3 moles of NaOH for complete reaction.

In this case, the number of moles of AI(NO3)3 (0.003 moles) is less than what is required for complete reaction with the given number of moles of NaOH (0.015 moles) in the same ratio.

Therefore, AI(NO3)3 is the limiting reactant, and NaOH is in excess.

To find out by how much NaOH is in excess, we need to determine the excess amount of NaOH.

Excess amount of NaOH = Total moles of NaOH - Moles of NaOH required for complete reaction

Total moles of NaOH = 0.015 moles
Moles of NaOH required for complete reaction = Moles of AI(NO3)3 * (3 moles of NaOH / 1 mole of AI(NO3)3)
= 0.003 moles * (3 moles of NaOH / 1 mole of AI(NO3)3)
= 0.009 moles

Excess amount of NaOH = 0.015 moles - 0.009 moles
= 0.006 moles

Finally, let's calculate the volume of excess NaOH in cm³.

Volume of excess NaOH = Excess amount of NaOH / Molarity of NaOH
= 0.006 moles / (0.15 moles/L)
= 0.04 L * (1000 cm³/ 1 L)
= 40 cm³

Therefore, the NaOH solution is in excess by 40 cm³.

So, the correct answer is option C. NaOH solution by 40cm³.