Answer:
1. 90%
2. 217.4 g O₂
3. 95.0%
4. Trial 2 ratios
Explanation:
Original: SiCl₄ + O₂ → SiO₂ + Cl₂
Balanced: SiCl₄ + O₂ → SiO₂ + 2Cl₂
Trial SiCl₄ O₂ SiO₂
1 120 g 240 g 38.2 g
2 75 g 50 g 25.2 g
<u>Percentage yield for trial 1</u>
We need to get actual yield (38.2 g) and theoretical yield, in grams.
Mass to moles:
molar mass SiCl₄: 28.09 + 4(35.45) = 169.9 g/mol
120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄
Moles to moles:
For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.
.706 mol SiCl₄ = .706 mol SiO₂
Moles to mass:
molar mass SiO₂: 28.09 + 2(16.00) = 60.09 g/mol
.706 mol SiO₂ x 60.09g/mol = 42.44 g SiO₂
Theoretical yield:
actual/theoretical x 100
38.2 / 42.44 = .900 = <u>90.0% yield</u>
<u>Leftover reactant for trial 1</u>
We know oxygen is the excess reactant.
Mass to moles:
molar mass O₂ = 32.00 g/mol
240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂
We used .706 mol SiO₂, so we also used .706 mol O₂.
7.5 - .706 = 6.8 moles left over
Moles to mass:
6.8 mol O₂ x 32.00g/mol =<u> 217.4 g O₂</u>
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<u>Percentage yield for trial 2</u>
Mass to moles:
molar mass SiCl₄: 169.9 g/mol
75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄
Moles to moles:
For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.
.441 mol SiCl₄ = .441 mol SiO₂
Moles to mass:
molar mass SiO₂: 60.09 g/mol
.441 mol SiO₂ x 60.09g/mol = 26.5 g SiO₂
Theoretical yield:
actual/theoretical x 100
25.2 / 26.5 = .950 = <u>95.0% yield</u>
Because the percentage yield of trial 2 is higher than that of trial 1, we know that the ratio of reactants in trial 2 is more efficient! We got a result closer to our theoretical yield.
