If a reaction mixture contains 3 moles of O2(g) with 4 moles of H2(g), which reactant is the limiting reactant and what is the mass (in grams) remaining of the reactant in excess?

Limiting reactant: H₂

Mass in grams remaining of the reactant in excess: 32 g

Explanation:

Reaction for the excersise is:  2H₂(g) + O₂(g) → 2H₂O (l)

Ratio is 2:1. We make these rules of three:

2 moles of hydrogen need to react with 1 mol of oxygen

Then 4 moles of H₂ will react with (4 .1) / 2 = 2 moles of O₂

We have 3 moles of O₂ and we need 2 moles, so the oxygen is my reagent in excess. 1 mol of O₂ remains (32 g) with no reaction.

Limiting reactant is the hydrogen, but let's verify:

1 mol of oxygen needs 2 moles of H₂ to react

Therefore, 3 moles of O₂ will react with (3 . 2) / 1 = 6 moles of H₂

We need 6 moles of H₂, but we have only 4 moles, not enough H₂

answer: mole of the gas

answer: mole of the gasexplanation:

5.6⋅10−^19  j

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