32.5k views
4 votes
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?

User Costi
by
4.4k points

1 Answer

1 vote

Answer:

94.1 %

Step-by-step 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 %

User Abzac
by
5.0k points