[student111]

For the reaction 2Al_(s) + 6H2O_(l) + 2NaOH_(aq) --> 2AL(OH)₄^- + 2Na_(aq)⁺ + 3H_{2 (g)}

you are given 49.92 grams of NaOH in a 0.6L solution and 41.28 grams of Al and you are asked to solve for the number of moles of H2 produced. I know it's a limiting reactant problem but the answer key says that NaOH is the limiting reactant. I am confused if the mole to mole ratio still holds for reactions with 3 reactants. For the equation above, how would you know that the hydrogens in NaOH lead to the H2? I think I may have over complicated the problem because I decided to split the beginning equation into two new equations:

3/2Al+6H₂O --> 3/2Al(OH)₄^- + 3H₂ (Equation 1)

1/2Al + 2NaOH --> 1/2Al(OH)₄^- + 2Na (Equation 2)

Both the equations are balanced and when you add these two equations, the sum is equal to the original equation. Based on this, I concluded that the NaOH wasn't directly producing the H2 gas, but the excess aluminum from the bottom reaction. I didn't worry about the water because I know there will be excess water, as 0.6 L = 600 grams = plenty. Using equation 2, I solved for the grams of aluminum from reacting 49.92 g of NaOH and I got 8.418 grams of Aluminum. Then I subtracted this value from the original 41.28 grams of Al in the problem to get the number of grams of aluminum that are left to react with the water. Then I solved for the moles of H2 using the first equation and I got a different answer than if I used the mole to mole ratio from the original equation. My guess for the cause of this discrepancy is that the reactions for the equations I made won't actually react. Basically, I am wondering whether or not my method works and why. And also if the mole to mole ratio works for reactions involving more than two reactants and products. I've attached my work below and sorry for the long read. Thanks!

[chenbeier]

Your Calculation is to complicated.

In the equation 2 Al correspond to 3 H₂

And 2 NaOH correspond to 3 H₂

So calculate the moles of each and find out how much hydrogen you get.

The lower number is the limiting agent.

[Borek]

Both the equations are balanced

No, they are not.

At first sight: in your first equation you are producing a negative charge out of nowhere.

I am confused if the mole to mole ratio still holds for reactions with 3 reactants.

It holds always, that's what whole stoichiometry is based on.

[student111]

The original equation also has a negative charge. I understand that the mole to mole ratio should hold true but why wouldn't my way work?

[chenbeier]

The original equation has negative and positive Charge.  Yours has only negative One. So it is not balanced.
You can only write Aluminium and water gives Aluminiumhydroxide and hydrogen.

2 Al + 6 H₂O => 2 Al(OH)₃ + 3 H₂

Then you add NaOH

Al(OH)₃ + NaOH => Na⁺ + [Al(OH)₄]^-

[Borek]

The original equation also has a negative charge. I understand that the mole to mole ratio should hold true but why wouldn't my way work?

Balanced equations has the same number of each element atoms on both sides atoms - that's just a mass conservation, atoms can't appear or disappear. Same about charge - it is conserved just like the mass is. There is a slight difference though - there can exist both positive and negative charge, so what matters is that the TOTAL charge doesn't change.

NaCl Na⁺ + Cl^-

is balanced, because total charge on the left is zero and total charge on the left is zero as well (+1 + (-1) = 0).

Na Na⁺

is not balanced, the charge on the right appears out of nowhere.

[MNIO]

I always recommend you follow these 6 steps in all stoichiometry problems (or a least some variation of them)
  (1) write balanced equations
  (2) convert everything given to moles
  (3) determine limiting reagent
  (4) convert moles limiting reagent to moles of other species
  (5) convert moles back to mass... this is the theoretical mass
  (6) % yield = (actual mass recovered / theoretical mass) * 100%

********
as to step (3), there are 3 common methods of determining the limiting reagent
   (method #1).. choose 1 reactant.  Determine the amount of the other reactant(s)
                         needed to complete consume the reactant you picked.  If you have
                         less than enough of the other reactant, the other reactant is the L.R.
                         If you have XS "other reactant" the reactant you chose originally is
                         the L.R.
  (method #2).. divide moles available by coefficients of the balanced equation for
                        each reactant the smallest ratio belongs to the limiting reagent
  (method #3).. convert all reactants to 1 product (steps 2, 4, and 5 combined).
                        Whichever gives the least amount of product is the L.R. and that
                        is the theoretical yield of that product.

you chose a rather unique variation of method #1.  Methods #2 or #3 would have worked much better.  And you didn't need to fret over the balanced equation.  Let me demonstrate

*********
first steps 1 and 2
    2Al(s) + 6H2O(l) + 2NaOH(aq) --> 2AL(OH)4- + 2Na(aq)+ + 3H2 (g)
    moles Al available = 41.28g * (1mol/26.982g) = 1.5298
    moles NaOH available = 49.92g  *(1mol/39.997g) = 1.2481
    moles H2O available ≈ 510 mL * (1g/1mL) * (1mol / 18.02g) = 28.3

method #1..
    I choose Al
       moles NaOH needed.  1.5298 mol Al * (2 mol NaOH / 2 mol Al) = 1.5298 mol NaOH
       moles H2O needed.  1.5298 mol Al * (6 mol H2O / 2 mol Al) = 4.5894 mol H2O
    since I have
       LESS moles of NaOH available than I need to completely consume all the Al
       MORE moles of H2O available than I need to completely consume all the Al
    therefore the NaOH is the limiting reagent.
   
method #2
   ratios of moles available to coefficients of balanced equation
       Al ratio = 1.5298 / 2 = 0.7649
       NaOH ratio = 1.2481 / 2 = 0.6240
       H2O ratio = 28.3 / 6 = 4.72
   since NaOH ratio < Al ratio < H2O ratio, the NaOH is the limiting reagent

both those methods require completing steps 4 and (we can skip 5) next
to finish the problem
    1.2481 mol Al * (3 mol H2 / 2 mol Al) = 1.872 mol H2

*******
method #3 is a bit different, we start with the balanced equation, then complete steps 2, and 4 (and 5 if needed) in 1 dimensional analysis equation for EACH reactant.  Whichever gives the least amount of H2 is the L.R and that is the theoretical yield.

and here's how that works... you start with the balanced equation

step (1)
   2Al(s) + 6H2O(l) + 2NaOH(aq) --> 2AL(OH)4- + 2Na(aq)+ + 3H2 (g)

then steps (2) and (4) combined.... (or 2+4+5 combined if mass is needed)
   IF Al is the LR
       41.28 g Al     1 mol Al       3 mol H2
       ----------- x ------------- x ----------- = 2.295 mol H2
              1          26.984g Al     2 mol Al

   IF NaOH is the LR
       49.92 g NaOH     1 mol NaOH          3 mol H2
       ---------------- x ----------------- x --------------- = 1.872 mol H2
                1              39.997g NaOH     2 mol NaOH

   IF H2O is the LR
       510 g H2PO      1 mol H2O        3 mol H2
       -------------- x -------------- x --------------- = 14.15 mol H2
               1             18.02g H2O     6 mol H2O

then step(3)
   since NaOH gave the least amount of H2, it is the LR
   and the theoretical yield of H2 = 1.872 mol

*********
*********
method #1 breaks down for determining LR for more than 2 reactants
method #2 works very well.. but it's not as compact as method #3...
method #3.. the holy grail of stoichiometric methods. . <---- your goal!!!