# How do I do stoichiometry for 2 limiting reactants

The Drano reaction produces sodium aluminate along with hydrogen gas. If a typical container of Drano weighs 500 g which is 35% by weight by weight sodium (l) hydroxide and 2.5% by weight aluminum, what volume hydrogen gas would be produced if (excess) water was added to an entire can of crystal Drano at 22 °C and 1.2 atm?

My equation:

$$\ce{4 Al + 4 Na(OH) + 12 H2O -> 4 Na[Al(OH)4] + 6 H2}$$

If my balancing correct, I am trying to get the excess for water. Would I need to do the moles left over? Or would 500 x 62.5% (2.5% and 35% added together) suffice?

If I do need to do moles left over, how do I manage with two limiting reactants?

• Richard, there can be only one limiting reactant. You cannot have two by definition. The one which is present in the least amount in a stoichiometric sense, is the limiting reactant! – M. Farooq Jun 21 at 1:14
• Besides the fact there is only one limiting reagent, the term "moles left over" is more applicable to the population of moles, the animals. In chemistry you probably want to use "extensive amount" instead. – andselisk Jun 21 at 4:53
• You can simplify the equation to $\ce{2Al + 2NaOH + 6H2O -> 2Na[Al(OH)4] + 3H2}$. – Mathew Mahindaratne Jun 21 at 16:30

## 2 Answers

Your balance equation is correct except for not simplifying it. Simplified equation should be: $$\ce{2Al + 2NaOH + 6H2O ⟶ 2Na[Al(OH)4] + 3H2}$$

I also think you're getting a little bit confused about your equation. You probably think reacting $$\ce{Al}$$ and $$\ce{NaOH}$$ with same mole ratio (2:2) make them two limiting reagents. If that is the case, you are absolutely wrong. As M. Farooq commented elsewhere, the one reactant, which is present in the least amount in a stoichiometric sense, is the limiting reactant. Let's see your question in hand:

The amount of $$\ce{NaOH}$$ in a typical container of Drano $$= \pu{500 g} \times \frac{\pu{35 g}\; \ce{NaOH}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{40 g}\; \ce{NaOH}} = \pu{4.38 mol}$$

The amount of $$\ce{Al}$$ in a typical container of Drano $$= \pu{500 g} \times \frac{\pu{2.5 g}\; \ce{Al}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{27 g}\; \ce{Al}} = \pu{0.463 mol}$$

Since $$\ce{NaOH}$$ and $$\ce{Al}$$ react with 1:1 ratio according to the equation, $$\ce{Al}$$ is clearly the limiting reagent.

Hence, $$\pu{0.463 mol}$$ of $$\ce{Al}$$ reacts with $$\pu{0.463 mol}$$ of $$\ce{NaOH}$$ if you added enough water to $$\pu{500 g}$$ of Drano container. The leftover amount of $$\ce{NaOH}$$ would be: $$\pu{(4.38 - 0.463) mol} = \pu{3.92 mol} = \pu{3.92 mol} \times \frac{\pu{40 g}\; \ce{NaOH}}{\pu{1 mol}} = \pu{157 g}$$

Now, how much water you need? You need more than $$\pu{0.463 mol}\times \frac{6}{2} = \pu{1.39 mol}$$ of water, which is more than $$\pu{25.0 g}$$ (to be excess).

Note: Similar calculations would appied to find out amount of $$\ce{H2}$$ released by the reaction.

I think you're getting a bit confused with your equation. It's probably better to write two distinct reactions, one with $$\ce{NaOH}$$ and one with $$\ce{Al}$$.

35% of $$\pu{500g}$$ is $$\pu{165g}$$ $$\ce{NaOH}$$, and 2.5% of $$\pu{500g}$$ is $$\pu{12.5g}$$ $$\ce{Al}$$.