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
2 s(s) + 3 o2(g) + 4 naoh(aq) → 2 na2so4(aq) + 2 h2o(l)

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× = 0,125 moles S

6,0g O₂× = 0,188 moles O₂

8,0g NaOH× = 0,2 moles NaOH

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

0,2 mol NaOH× = 0,1 moles of Sulfur

0,2 mol NaOH× = 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!

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.

bcde is the incorrect answer it is a