Question

In the following reaction, oxygen is the excess reactant.
SiCl4 + O2 → SiO2 + Cl2
The table shows an experimental record for the above reaction.
trail 1: Starting Amount of SiCl4: 150 g Starting Amount of O2: 200 g Actual Yield of SiO2: 49.2 g
trail 2: Starting Amount of SiCl4: 75 g Starting Amount of O2:50 g Actual Yield of SiO2: 25.2 g
a. Calculate the percentage yield for SiO2 for Trial 1. Also, determine the leftover reactant for the trial. Show your work.
b. Based on the percentage yield in Trial 2, explain what ratio of reactants is more efficient for the given reaction.

Answer 1

The equilibrium constant for the given reaction is 100. <br> `N_(2)(g)+2O_(2)(g) hArr 2NO_(2)(g)` <br> What is the equilibrium constant for the reaction ? <br> `NO_(2)(g) hArr 1//2 N_(2)(g) +O_(2)(g)`

Answer 2

Answer:

a) the percentage yield of the experiment is 92.13% and there is 171.52 grams of Oxygen as leftover reactant.

b) the percentage yield of Trial 2 is higher, the ratio of reactants 3:2 is more efficient.

Explanation:

The balanced reaction will be as follows;

$SiCl_{4} + O_{2} ------ > SiO_{2} + 2Cl_{2}$

Here one mole of $SiCl_{4}$ reacts with one mole of $O_{2}$ to form one mole of $SiO_{2}$ and two moles of $Cl_{2}$.

Molar mass of $SiCl_{4} = 168 g$

Molar mass of $O_{2} = 32 g$

Molar mass of $SiO_{2} = 60 g$

Molar mass of $Cl_{2} = 35 g$

$Percentage Yield = \frac{Actual Yield}{Theoretical Yield}*100$

Trial 1:

Starting amount of $SiCl_{4}$ is $150 g$ which is,

$\frac{150 g}{168 g} = 0.89 moles$

So, we can conclude 0.89 mole of $SiCl_{4}$ reacts with 0.89 mole of $O_{2}$ to form 0.89 mole of $SiO_{2}$ and 1.78 moles of $Cl_{2}$.

Theoretical yield of $SiO_{2} = 0.89 mol * 60 gmol^{-1} = 53.4g$

Actual yield of $SiO_{2} = 49.2 g$

$Percentage Yield = \frac{49.2 g}{53.4 g}*100 = 92.13 percent$

Thus, the percentage yield of the experiment is 92.13%.

So, leftover reactant,

$O_{2} = 200 g - (32*0.89) g = 171.52 g$

Thus, the leftover reactant is 171.52 grams of Oxygen.

$Ratio of Reactants SiCl_{4}:O_{2} = \frac{150 g}{200 g}=3:4$

Trial 2:

Starting amount of $O_{2}$ is $50 g$ which is,

$\frac{50 g}{32 g} = 1.56 moles$

Starting amount of $SiCl_{4}$ is $75 g$ which is,

$\frac{75 g}{168 g} = 0.45 moles$

So, we can conclude 0.45 mole of $SiCl_{4}$ reacts with 0.45 mole of $O_{2}$ to form 0.45 mole of $SiO_{2}$ and 0.45 moles of $Cl_{2}$.

Theoretical yield of $SiO_{2} = 0.45 mol * 60 gmol^{-1} = 27g$

Actual yield of $SiO_{2} = 25.2 g$

$Percentage Yield = \frac{25.2 g}{27g}*100 = 93.33 percent$

Thus, the percentage yield of the experiment is 93.33%.

$Ratio of Reactants SiCl_{4}:O_{2} = \frac{75 g}{50 g}=3:2$

Since, the percentage yield of Trial 2 is higher, the ratio of reactants 3:2 is more efficient.