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:

b) Mole ratio:

c) Limiting reactant:

d) Convert 100 g of SiCl₄ to moles

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

     ⇒ x = 0.589 mol SiO₂

f) Convert 0.589 mol SiO₂ to grams:

g) Compute percent yield:

3) Leftover reactant

a) Convert 100 g of O₂ to moles

b) Excess:

c) Convert 2.54 mol of O₂ to grams:

4) Percent yield for trial 2

a) Convert 75 g of SiCl₄ to moles

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

     ⇒ x = 0.441 mol SiO₂

c) Convert 0.441 mol SiO₂ to grams:

d) Compute the percentage yield:

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.

the temperature of the gas, in k, not celsius.

it means there are 3 atoms of oxygen