If 425.0 grams of butene react completely in excess oxygen, how many grams of each product are produced? How many grams would be produced if the percent yield is 67? At STP what would be the volume of the gas produced for this yield?

Answers:

A. 1333 g CO₂; 546.0 g H₂O

B. 2200 g CO₂; 370 g H₂O

C. 460 L

Step-by-step explanation:

We know we will need a balanced equation with masses and molar masses, so let's gather all the information in one place.

M_r: 56.11 44.01 18.02

C₄H₈ + 6O₂ ⟶ 4CO₂ + 4H₂O

m/g: 425.0

A. Theoretical yield of each product

(1) Calculate the moles of C₄H₈

n = 425.0 * 1/56.11

n = 7.574 mol C₄H₈

(2) Calculate the moles of CO₂ and H₂O

In each case, the molar ratio is 4 mol product/1 mol C₄H₈.

CO₂ and H₂O:

n = 7.574 * 4/1

n = 30.30 mol

(3) Calculate the masses of CO₂ and H₂O

Mass of CO₂ = 30.30 mol CO₂ * (44.01 g CO₂/1 mol CO₂)

Mass of CO₂ = 1333 g CO₂

Mass of H₂O = 30.30 mol H₂O * (44.01 g H₂O/1 mol H₂O)

Mass of H₂O = 546.0 g H₂O

B. Actual yield of each product

Mass of CO₂ = 3333 g * 67/100

Mass of CO₂ = 2200 g CO₂

Mass of H₂O = 546.0 g * 67/100

Mass of H₂O = 370 g H₂O

C. Volume of gas

At STP, CO₂ is a gas, but H₂O is a liquid.

Moles of gas = moles of CO₂

If the actual yield is 67 %

Moles of CO₂ = 30.30 mol * 67/100

Moles of CO₂ = 20 mol

STP is 1 bar and 0 °C.

The molar volume at STP is 22.71 L.

∴ V = 201 mol * 22.71 L/1 mol

V = 460 L