Question

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:
`2S(s) + 3O_2(g) + 4NaOH(aq) \rightarrow 2Na_2SO_4(aq) + 2H_2O(l)`a. O₂(g)b. NaOH(aq)c. S(s)d. None of these substances is the limiting reactant.

Answer 1

The limiting reactant is NaOH (option B)

Step-by-step explanation:

2S(s) + 3O₂(g) + 4NaOH(aq) → 2Na₂SO₄(aq) + 2H₂O(l)

The reaction is ballanced. OK

We need to know how many moles do we have from each compound.

Mass / Molar weight = Mol

Molar weight S = 32 g/m

Molar weight O₂ = 32 g/m

Molar weight NaOH = 40 g/m

Mol S: 2g/ 32g/m = 0.0625 mol

Mol O₂: 3g / 32 g/m = 0.09375 mol

Mol NaOH: 4g/ 40g/m = 0.1 mol

Now, we can play with the reactants. The base is: 2 moles of S, react with 3 mol of O₂ and 4 moles of hydroxide to make 2 moles of sulfate and 2 moles of water. Pay attention to the rules of three.

2 moles of S __ react with __ 3 moles of O₂ __ and __ 4 moles of NaOH

0.0625 moles S __________ 0.09375 moles O₂ ___ 0.125 moles NaOH

The limiting reactant is the NaOH. I need to use 0.125 moles and I only have 0.1 moles.

Let's do the same with O₂

3 moles of O₂ __ react with __ 2 moles of S __ and __ 4 moles of NaOH

0.09375 moles of O₂ _______ 0.0625 mol of S _____ 0.125 moles NaOH