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)

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:

S: 2 moles
O₂: 3 moles
NaOH: 4 moles
Na₂SO₄: 2 moles
H₂O: 2 moles

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

S: 32 g/mole
O: 16 g/mole
Na: 23 g/mole
H: 1 g/mole

Therefore, the molar mass of each compound is:

S: 32 g/mole
O₂: 2* 16 g/mole= 32 g/mole
NaOH: 23 g/mole+16 g/mole+ 1 g/mole= 40 g/mole
Na₂SO₄: 2*23 g/mole + 32 g/mole + 4*16 g/mole= 142 g/mole
H₂O: 2* 1 g/mole + 16 g/mole= 18 g/mole

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

S: 2 moles* 32 g/mole= 64 g
O₂: 3 moles* 32 g/mole= 96 g
NaOH: 4 moles* 40 g/mole= 160 g
Na₂SO₄: 2 moles* 142 g/mole= 284 g
H₂O: 2 moles* 18 g/mole= 36 g

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?

[tex]mass of O_{2} =\frac{2 grams of S*96 grams of O_{2} }{64 grams of S}[/tex]

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?

[tex]mass of NaOH=\frac{2 grams of S*160 grams of NaOH }{64 grams of S}[/tex]

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.