Question

Which substance is the limiting reactant when 34 g of sulfur reacts with 54 g of oxygen and 68 g of potassium hydroxide according to the following chemical equation? 2 s(s) + 3 O₂(g) + 4 KOH(aq) → 2 K₂SO₄(aq) + 2 H₂O(l)

Answer 1

To find the limiting reactant, we calculate the moles of each reactant provided and compare them with the stoichiometry of the reaction. Potassium hydroxide (KOH) is the limiting reactant because it has the fewest moles compared to what is required by the balanced equation.

Step-by-step explanation:

The question is asking to determine the limiting reactant in a chemical reaction where sulfur reacts with oxygen and potassium hydroxide, according to the given balanced chemical equation.

To find the limiting reactant, we need to calculate the number of moles of each reactant and then use the stoichiometry of the balanced equation to determine which reactant will run out first. For sulfur (S), with an atomic mass of approximately 32 g/mol, 34 g corresponds to roughly 34 g / 32 g/mol = 1.0625 moles. For oxygen (O₂), using a molar mass of 32 g/mol, 54 g corresponds to 54 g / 32 g/mol = 1.6875 moles. For potassium hydroxide (KOH), its molar mass is approximately 56 g/mol, so 68 g corresponds to 68 g / 56 g/mol = 1.2143 moles.

Referring to the balanced chemical equation, 2 moles of sulfur react with 3 moles of oxygen and 4 moles of KOH. Thus, for the moles of sulfur we have (1.0625 moles S x (3 moles O₂/2 moles S)) = 1.59375 moles of O₂ required and (1.0625 moles S x (4 moles KOH/2 moles S)) = 2.125 moles of KOH required. As we only have 1.6875 moles of O₂ and 1.2143 moles of KOH, it becomes clear that KOH is the limiting reactant because we have less than 2.125 moles that are required.