Question

HELP!!!

Answer 1

Answer:

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 mass

• molar mass of SiCl₄: 169.9 g/mol

• moles = 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 mass

• molar mass SiO₂ = 60.08 g/mol

• theoretical yield of SiO₂ = 0.589 mol × 60.08 g/mol = 35.4 g

g) Compute percent yield:

• percent yield = (actual yield / theoreticl yield) × 100

• percent 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 mass

• molar mass of O₂ = 32.0 g/mol

• moles 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 mass

• molar mass of SiCl₄: 169.9 g/mol

• moles = 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 mass

• molar mass SiO₂ = 60.08 g/mol

• theoretical yield of SiO₂ = 0.441 mol × 60.08 g/mol = 26.5 g

d) Compute the percentage yield:

• percentage yield = (actual yield / theoreticl yield) × 100

• percentage 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.