in the following reaction, oxygen is the excess reactant. sicl4 + o2 → sio2 + cl2 the table shows an experimental record for the above reaction. experimental record trial starting amount of sicl4 starting amount of o2 actual yield of sio2 1 100 g 100 g 32.96 g 2 75 g 50 g 25.2 g calculate the percentage yield for sio2 for trial 1. also, determine the leftover reactant for the trial. show your work. based on the percentage yield in trial 2, explain what ratio of reactants is more efficient for the given reaction.

1. Percent yield Trial 1: 93.1 %

2. Leftover reactant for Trial 1: 81.3 g

3. Percent yield Trial 2: 95.1%

4. What ratio of reactants is more efficient: the ratio of the Trial 2: 75g / 50g

Explanation

1) Data (the table is copied for better understanding of the data)

Trial     Start. Amount of SiCl₄    Start, Amount of O₂   Actual Yield of SiO₂

  1           100 g                                100 g                            32.96 g

  2           75 g                                  50 g                              25.2 g

2) Percent yield for SiO₂ for Trial 1.

a) Balanced chemical equation:

SiCl₄ + O₂ → SiO₂ + 2Cl₂

b) Mole ratio:

1 mol SiCl₄ : 1 mol O₂ : 1 mol SiO₂ : 2Cl₂

c) Limiting reactant:

SiCl₄ (it is given that oxygen is the excess reactant)

d) Convert 100 g of SiCl₄ to moles

moles = mass in grams / molar massmolar mass of SiCl₄: 169.9 g/molmoles = 100 g / 169.9 g/mol = 0.589 mol

e) Find moles of SiO₂ using proportions and theoretical mole ratio:

1 mol SiCl₄ /  1 mol SiO₂ = 0.589 mol SiCl₄ / x

     ⇒ x = 0.589 mol SiO₂

f) Convert 0.589 mol SiO₂ to grams:

mass in grams = number of moles × molar massmolar mass SiO₂ = 60.08 g/moltheoretical yield of SiO₂ = 0.589 mol × 60.08 g/mol = 35.4 g

g) Compute percent yield:

percent yield = (actual yield / theoreticl yield) × 100percent yield = (32.96 g / 35.4 g) × 100 = 93.1%

3) Leftover reactant

a) Convert 100 g of O₂ to moles

moles = mass in grams / molar massmolar mass of O₂ = 32.0 g/molmoles of O₂ = 100 g / 32.0 g/mol = 3.13 mol

b) Excess:

3.13 mol - 0.589 mol = 2.54 mol O₂

c) Convert 2.54 mol of O₂ to grams:

mass = number of moles × molar mass = 2.54 mol × 32.0 g/mol = 81.2 g

4) Percent yield for trial 2

a) Convert 75 g of SiCl₄ to moles

moles = mass in grams / molar massmolar mass of SiCl₄: 169.9 g/molmoles = 75 g / 169.9 g/mol = 0.441 mol

b) Find moles of SiO₂ using proportions and theoretical mole ratio:

1 mol SiCl₄ /  1 mol SiO₂ = 0.441 mol SiCl₄ / x

     ⇒ x = 0.441 mol SiO₂

c) Convert 0.441 mol SiO₂ to grams:

mass in grams = number of moles × molar massmolar mass SiO₂ = 60.08 g/moltheoretical yield of SiO₂ = 0.441 mol × 60.08 g/mol = 26.5 g

d) Compute the percentage yield:

percentage yield = (actual yield / theoreticl yield) × 100percentage yield = (25.2 g / 26.5 g) × 100 = 95.1%

e) Conclusion: Since the second trial had a greater percentage yield than the first trial (95.1% vs 93.1 %), the ratio of the Trial 1 is more efficient for the given reaction.

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