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Which substance is the limiting reactant when 2.0 g of sulfur reacts with 3.0 g of oxygen and 4.0 g of sodium hydroxide according to the following chemical equation? 2 S(s) + 3 O2(g) + 4 NaOH(aq) → 2 Na2SO4(aq) + 2 H2O(l)

Sep 24, 2015

Sodium hydroxide.

Explanation:

Don't be confused by the fact that you have three reactants, you can find the limiting reagent by using the same technique you use for reactions that have two reactants.

So, start by taking a look at the balanced chemical equation

$\textcolor{red}{2} {\text{S"_text((s]) + color(blue)(3)"O"_text(2(g]) + color(green)(4)"NaOH"_text((aq]) -> 2"Na"_2"SO"_text(4(aq]) + 2"H"_2"O}}_{\textrm{\left(l\right]}}$

Notice that mole ratios that exist between the three reactants. In order for the reaction to take place, you need $\textcolor{red}{2}$ moles of sulfur for every $\textcolor{b l u e}{3}$ moles of oxygen gas and $\textcolor{g r e e n}{4}$ moles of sodium hydroxide.

If the numbers of moles of each reactant do no not follow this ratio, then you will indeed have a limiting reagent.

Use the molar mass of each reactant to find how many moles of each you have

2.0color(red)(cancel(color(black)("g S"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g S")))) = "0.0624 moles S"

3.0color(red)(cancel(color(black)("g O"_2))) * ("1 mole O"""_2)/(32.0color(red)(cancel(color(black)("g O"_2)))) = "0.0938 moles O"""_2

4.0color(red)(cancel(color(black)("g NaOH"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g NaOH")))) = "0.100 moles NaOH"

Now, let's say that we want to compare the number of moles of oxygen gas and sodium hydroxide with the number of moles of sulfur. Remember, you want to check if the amounts of reactants you have satisfy those mole ratios.

0.0624color(red)(cancel(color(black)("moles S"))) * (color(blue)(3)" moles O"""_2)/(color(red)(2)" moles S") = "0.0936 moles O"""_2

So, do you have enough oxygen gas?

Yes, you do, since you need 0.0936 moles and have 0.0938 moles. The oxygen gas will not be the limiting reagent.

Do the same for sodium hydroxide.

0.0624color(red)(cancel(color(black)("moles S"))) * (color(green)(4)" moles NaOH")/(color(red)(2)" moles S") = "0.125 moles NaOH"

Do you have enough sodium hydroxide?

No, you don't. You need 0.125 moles, but only have 0.100 moles. This means that sodium hydroxide will be the limiting reagent.

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