Calculate the percentage yield for SiO2

Question

In the following reaction, oxygen is the excess reactant.

SiCl₄ + O₂ → SiO₂ + Cl₂

The table shows an experimental record for the above reaction.

Experimental RecordTrialStarting Amount of SiCl4Starting Amount of O2Actual Yield of SiO2
1	100 g	100 g	32.96 g
2	75 g	50 g	25.2 g

1. Calculate the percentage yield for SiO₂ for Trial 1. Also, determine the leftover reactant for the trial. Show your work.
2. Based on the percentage yield in Trial 2, explain what ratio of reactants is more efficient for the given reaction.

Answer 1

I'll assume the table looks like this:

Balance the chemical reaction equation:

SiCl₄ + O₂ → SiO₂ + 2Cl₂

Turn 100 g of SiCl4 into moles by first calculating the molar mass:

Si: 28.086 x 1 = 28.086

Cl: 35.453 x 4 = 141.812

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SiCl₄: = 169.898 g/mol

(100 g)/(169.898 g/mol) = 0.5886 moles

Using the chemical reaction equation, 1 mole of SiCl₄ produces 1 mole of SiO₂ therefore there are 0.5886 moles of SiO₂ produced in a 100% yield.

Calculate the grams of SiO₂ produced by first calculating the molar mass of SiO₂:

Si: 28.086 x 1 = 28.086

O: 15.999 x 2 = 31.998

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SiO₂: = 60.084 g/mol

(0.5886 moles)(60.084 g/mol) = 35.36 g of SiO₂

Calculate the percent yield:

32.96/35.36 x 100 = 93.2%

Calculate the left over reactant:

Per the chemical reaction equation, 1 mole of O₂ is required along with 1 mole of SiCl₄ for the reaction therefore 0.5886 moles of O₂ was required.

The molar mass of O₂ is 15.999 x 2 = 31.998 g/mole so the amount of O₂ consumed in the reaction was (0.5886 moles)(31.998 g/mol) = 18.83 g. Since we started with 100 g, there were 81.17 g of O₂ left over

Use the same process to find the % yield for trial 2 and compare the answer to the 93.2% you got for trial 1. Whichever is highest is the "more efficient" ratio of reactants.