Answer:
Sodium hydroxide NaOH will be the limiting reagent.
Step-by-step 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:
- 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?
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.