If you start with 13 mole H2O and 50 grams of O2 what is the percent yield of H202 if 100 grams of H202 is actually made?

Chemistry

1 answer:

dmitriy555 [2]3 years ago

3 0

Answer:

94.1 %

Explanation:

We firstly determine the equation:

2H₂O + O₂ → 2H₂O₂

2 moles of water react to 1 mol of oxygen in order to produce 2 moles of oxygen peroxide.

We convert the mass of oxygen to moles:50 g . 1mol /32g = 1.56 mol

Certainly oxygen is the limiting reactant.

2 moles of water react to 1 mol of oxygen.

13 moles of water may react to 13/2 = 6.5 moles. (And we only have 1.56)

As we determine the limiting reactant we continue to the products:

1 mol of O₂ can produce 2 moles of H₂O₂

Then 1.56 moles of O₂ will produce (1.56 . 2) = 3.125 moles

We convert the moles to mass: 3.125 mol . 34 g/mol= 106.25 g

That's the 100% yield or it can be called theoretical yield.

Percent yield = (Yield produced / Theoretical yield) . 100

(100g / 106.25 g) . 100 = 94.1 %

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<h3>Explanation</h3>

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How many moles of thiosulfate ions ${\text{S}_2\text{O}_3}^{2-}$ are required?

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