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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

2 answers

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?