Stoichiometry Problem

Question

 If 50.0 grams of silicon dioxide is heated with excess of carbon, 27.9 grams of silicon carbide is produced during the experiment.

                                SiO₂(s) + 3C(s) → SiC(s) + 2CO

             What is the percent yield of this reaction?

Answer 1

Hello, Lilia,

The chemical equation is balanced, and we are told that there is excess carbon. We can conclude that all 27,9 grams of the SiO₂ reacts to form the silicon carbide, SiC. The equation states that we should get 1 mole of SiC for every 1 mole of SiO₂, a molar ratio of 1/1.

That means if we know the moles of SiO₂, we will also know the moles of SiO to expect. 50.0 grams of SiO₂, divided by the molar mass of SiO₂, will give us the moles: 50.0g/60.1g/mole = 0.832 moles SiO₂. We expect the same moles of SiC, 0.832 moles SiC. Multiply that times the molar mass of SiC to obtain grams SiC: I get 33.4 grams.

Actual yield was 27,9 grams of SiC. The percent yield is therefore (27.9g/33.4g)x100% = 83.4% (3 sig figs).

Not bad, but such a result would normally require some discussion on what may have gone wrong and what changes might be made to improve the yield. It is good to have a random number generator to pick from an established set of time-tested excuses:

1. My labmate dropped the beaker
2. My dog ate some
3. It looks like silver, what did you expect?
4. Just kidding

Bob

Answer 2

percent yield = (actual yield / theoretical yield) * 100%

Theoretical yield refers to what you calculate and is the maximum product you can get

Actual yield refers to the grams of product (for this problem this value is 27.9 g of SiC)

We are also given a balanced chemical equation:

SiO₂(s) + 3C(s) → SiC(s) + 2CO

SiO₂: 50 g

SiC: X g (theoretical) & 27.9 (actual)

First we must calculate the theoretical yield by converting our reactant SiO₂ from grams to moles using its molar mass.

50 g SiO₂ * (1 mol SiO₂ / 60.08 g SiO₂) = 0.832 mol SiO₂

Now we must convert moles of reactant (SiO₂) to moles of product (SiC) using the coefficients from the chemical reaction. For 1 mole of SiO_2, there is 1 mole of SiC.

0.832 mol SiO₂ * (1 mol SiC / 1 mol SiO₂) = 0.832 mol SiC

Convert moles of SiC to grams using its molar mass

0.832 mol SiC * (40.11 g SiC / 1 mol SiC) = 33.38 g SiC (this is our theoretical yield, X)

We can substitute our values into this formula:

percent yield = (actual yield / theoretical yield) * 100%

(27.9 g / 33.38 g) * 100% = 83.6% percent yield

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