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.) 120 g 240 g 38.2 g

2.) 75 g 50 g 25.2 g

b

Explanation:

Trial 1

Percent yield = 90.09 %

leftover reactant(O2) = 217.408 g

Trial 2

percent yield = 95.94 %

leftover reactant(O2) = 35.48 g

The most efficient proportion is 75 of SiCl4 : 50 g of O2 because from this proportion a higher percentage of yield was obtained

Explanation

we calculate percent yield from of equation

actual yield

percent yield = x100

theoretical yield

we calculate theoretical yield by stoichiometry

first we calculate limiting reagent

For Trial 1  

mass

mol SiCl4=

Mass Molar

Mass Molar = 169.86 g/mol

120 g

mol SiCl4=

169.86 g/mol

mol SiCl4= 0.706 mol

240 g

mol O2=

32 g/mol

mol O2= 7.5 mol

assuming that all SiCl4 reacts completely we must calculate the O2 moles needed for all siCl4 to react completely  

SiCl4  + O2 → SiO2 + 2Cl2

0.706 mol X

0.706 mol SiCl4 x 1 mol O2

mol O2=

1 mol SiCl4

mol O2 = 0.706 The amount of oxygen needed for all SiCl4 to react is lower than there is initially oxygen so that oxygen is overreactive and SiCl4 is the reactive limit  

Now from of limiting reagent we calculate mol of SiO2

SiCl4  + O2 → SiO2 + 2Cl2

0.706 mol X

0.706 mol SiCl4 x 1 mol O2

mol SiO2=

1 mol SiCl4

mol SiO2= 0.706 mol this mol we transformed to grams as follow

Mass Molar= 60.06 g/mol

mass SiO2= 0.706 mol x 60.06 g/mol

mass SiO2= 42.40 g

Then percent yield is

38.2 g

Percent yield = x100

42.40 g

percent yield = 90.09 %

Regent in excess is O2 we can calculate mol in excess as follow

mol in excess O2 = initial mol - mol of reaction

mol in excess= 7.5 mol -0.706

mol in excess= 6.794 mol

mass = mol x Mass Molar

mass in excess= 6.794 mol x 32g/mol

mass In excess= 217.408 g

Explanation:

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