Stoichiometry Problems!

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C₆H₁₂O₆(s) + 6O₂(g) => 6CO₂(g) + 6H₂0(l)

a.)

First we need to find how many molecules of oxygen gas we need.

Every one molecule of C₆H₁₂O₆ will react with six molecules of O₂. So, if we have 3.011 x 10²³ molecules of C₆H₁₂O₆, we need six times that of oxygen.

3.011 x 10²³ x 6 = 18.066 x 10²³ = 1.8066 x 10²⁴

So we need 1.8066 x 10²⁴ molecules of O₂. We need to find the volume of this in liters.

At STP, one mole of a gas occupies 22.4 liters. Let's find the number of moles we have of O₂.

(1.8066 x 10²⁴) ÷ (6.022 x 10²³) = 3 moles

3 x 22.4 = 67.2

67.2 liters of O₂ is needed.

b.)

Okay, so to find the percent yield, we need to find the theoretical yield and the actual yield. We are given the actual yield, so what we need is the theoretical yield.

For every one mole of C₆H₁₂O₆, theoretically 6 moles of H₂O will be produced.

Let's convert grams to moles for C₆H₁₂O₆:

1 gram / 180 grams = 0.0055556 moles C₆H₁₂O₆

Theoretically, 6 times that is the moles of H₂O produced:

0.0055556 x 6 = 0.033333 moles H₂O

Molar mass of H₂O is 18.015, so let's find grams:

0.033333 x 18.015 = 0.600 grams H₂O

So we have our theoretical yield, 0.600, and our actual yield, 0.303.

0.303 ÷ 0.600 = 0.505

Convert to a percent: 0.505 x 100 = 50.5%

The percent yield is 50.5%.

Hope this helps!

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