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PLSSSS HELPPPP !!!!!

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.

User M Rostami
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1 Answer

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To calculate the percentage yield for Trial 1, we need to use the actual yield and theoretical yield of SiO2. The theoretical yield can be calculated using the balanced chemical equation and the starting amount of limiting reactant (SiCI4 in this case).

The balanced chemical equation is: SiCI4 + 02 → SiO2 + C12

From the experimental record, we know that the starting amount of SiCI4 for Trial 1 is 1120 g. We can use this information to find the theoretical yield of SiO2:

Molar mass of SiCI4 = 169.9 g/mol
Molar mass of SiO2 = 60.1 g/mol

Number of moles of SiCI4 = 1120 g / 169.9 g/mol = 6.59 mol
Number of moles of SiO2 produced (from stoichiometry) = 6.59 mol x (1 mol SiO2 / 1 mol SiCI4) = 6.59 mol

Theoretical yield of SiO2 = 6.59 mol x 60.1 g/mol = 395.5 g

Now we can calculate the percentage yield for Trial 1:

Percentage yield = (actual yield / theoretical yield) x 100%
Percentage yield = (38.2 g / 395.5 g) x 100%
Percentage yield = 9.65%

To determine the leftover reactant for Trial 1, we need to use the balanced chemical equation and the amount of excess reactant (oxygen in this case).

Number of moles of oxygen used = 240 g / 32 g/mol = 7.5 mol (from Trial 1)
Number of moles of SiCI4 used (from stoichiometry) = 7.5 mol x (1 mol SiCI4 / 1 mol O2) = 7.5 mol
Starting moles of SiCI4 = 1120 g / 169.9 g/mol = 6.59 mol
Moles of SiCI4 left = 6.59 mol - 7.5 mol = -0.91 mol

The negative value for moles of SiCI4 left indicates that SiCI4 is in excess and there is no leftover reactant of oxygen.

From the percentage yield in Trial 2 (31.8%), we can see that the ratio of reactants used in Trial 2 was more efficient than in Trial 1. To determine the optimal ratio, we would need to calculate the theoretical yield for various ratios of reactants and compare their percentage yields. However, we can make a general observation that the ratio of reactants in Trial 2 produced a higher percentage yield than in Trial 1, so it may be more efficient.
User LightMan
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