Question

Which substance is the limiting reactant when 4.0 g of sulfur reacts with 6.0 g of oxygen and 8.0 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)

Answer 1

Answer:

Limiting reactant is NaOH

Explanation:

In the reaction:

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

4.0 g of sulfur, 6.0 g of oxygen and 8.0 g of sodium hydroxide are:

4,0g S×$\frac{1mol}{32,065g}$ = 0,125 moles S

6,0g O₂×$\frac{1mol}{32g}$ = 0,188 moles O₂

8,0g NaOH×$\frac{1mol}{40g}$ = 0,2 moles NaOH

For a complete reaction of 0,2 moles of NaOH you will need:

0,2 mol NaOH×$\frac{2molS}{4molNaOH}$ = 0,1 moles of Sulfur

0,2 mol NaOH×$\frac{3molO_{2}}{4molNaOH}$ = 0,15 moles of Oxygen

As you have 0,125 moles of S and 0,188 moles of O₂ that are more than moles you need for a complete reaction of NaOH, limiting reactant is NaOH

I hope it helps!