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.