Question

How do I identify the limiting and excess reagents, and how much excess is present for AlCl3 (aq) + 3NaOH (aq) -> AlNa3 + 3ClOH? Also, how do I know the products' states?

Answer 1

AlCl3 (aq) + 3NaOH (aq) -> AlNa3 (s) + 3ClOH
20 g 19 g
133.33 g/mol 40 g/mol

This is what I did:

\[nAlCl3 = 20.0g \times 1mol \div 133.33 g = 0.15 mol\]

mole ratio AlCl3:NaOH = 1:3

\[nNaOH needed = 0.15 mol AlCl3 \times 3 mol O2 \div 1 mol AlCl3 = 0.45 mol O2\]

\[nNaOH available = 19.0 g \times 1 mol \div 40 g = 0.475 mol\]

I'm so confused because the correct answer is supposed to be 0.025 mol excess NaOH (aq). Why? Someone?

Answer 2

Did you figure what the limiting reagent is?

Answer 3

no, I did not.. how do I figure it out?

Answer 4

Well, you were on the right track, but I guess you got a bit confused.

So first find the moles of each reactant.
AlCl3: 20 g / 133.33 g/mol = 0.15 mol
NaOH: 19 g / 40 g/mol = 0.475 mol

Now if we had an unlimited amount of NaOH and only 0.15 mol of AlCl3, 0.15 mol of AlNa3 would be produced.
If we had an unlimited amount of AlCl3 and only 0.475 mol of NaOH, 0.158 mol of AlNa3 would be produced.

So AlCl3 is the limiting reagent in this case. It will run out before all of the NaOH runs out.

Answer 5

Now since AlCl3 is the limiting reagent, all 0.15 moles of it will be used up to produce 0.15 moles of AlNa3.
0.45 mol of NaOH would be used up in producing that 0.15 mol of AlNa3. The remaining NaOH is 0.475 mol - 0.45 mol = 0.025 mol.

Answer 6

I don't understand:

"Now if we had an unlimited amount of NaOH and only 0.15 mol of AlCl3, 0.15 mol of AlNa3 would be produced. If we had an unlimited amount of AlCl3 and only 0.475 mol of NaOH, 0.158 mol of AlNa3 would be produced." this part D:

Answer 7

Well, you can disregard the "if we had an unlimited amount of..." the point is that you can only produce 0.15 mol of a certain product (AlNa3) with 0.15 mol of AlCl3. With the amount of NaOH, you can produce more of the same product. When the reaction takes place with 0.15 mol of AlCl3 and 0.475 mol of NaOH, the AlCl3 will run out first.

Answer 8

Do you get it or am I still not really explaining it?

Answer 9

ohh, i see. I'm getting it-- thank you! :)

Answer 10

Alright, great! =)

Answer 11

you've got the wrong products, so the whole answer above is wrong.

Answer 12

really?

Answer 13

AlNa3 is not a proper formula, since both Al and Na will form positive ions. Cl and OH will both form negative ions, so they won't pair up either. The proper reaction is\[AlCl{_3} + 3NaOH \rightarrow 3NaCl + Al(OH){_3}\]

Answer 14

ohh, i see. But the calculations are still the same-- like the way to do it?

Answer 15

if all you're doing is comparing the reactants, then yes, the approach is still the same