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)

Sodium hydroxide NaOH will be the limiting reagent.

Explanation:

By stocheimetry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:

On the other hand, you know the following masses of each element that make up the compounds:

Therefore, the molar mass of each compound is:

Therefore, by stoichiometry of the reaction, the following amounts of the compounds react by mass or are produced:

Now, you apply a simple rule of three as follows: if 64 grams of S reacts with stoichiometry with 96 grams of O₂, 2 grams of S with how much mass of O₂ will react?

mass of O₂= 3 grams

You have enough amount of O₂ since it needs 3 grams and has 3 grams. The oxygen gas will not be the limiting reagent.

Now, you apply the following rule of three: if 64 grams of S reacts with stoichiometry with 160 grams of NaOH, 2 grams of S with how much mass of NaOH will react?

mass of NaOH= 5 grams

In this case you need 5 grams of NaOH, but it only has 4 grams. This indicates that you do not have enough, which means that sodium hydroxide will be the limiting reagent.