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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
1 120 g 240 g 38.2 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. 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

Percentage yield for trial 1

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 = 90.0% yield

Leftover reactant for trial 1

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 = 217.4 g O₂

Percentage yield for trial 2

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 = 95.0% yield

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.

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it

the answer is definitely c.