Trial: Starting Amount of SiCl4: 150g Starting Amount of O2:200g Actual Yield of SiO2:49.2 g trial 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.

a. The percentage yield of SiO₂ for Trial 1 is approximately 93.31%

The leftover reactants are 10.8892 g of SiCl₄ and 173.794944 g of O₂

b. The ratio of the reactants, SiCl₄ to O₂, in Trial 2 of 3:2 is more efficient than that of Trial 1, which is 3:4

Explanation:

a. The chemical reaction can be expressed as follows;

SiCl₄ (g) + O₂ (g) → SiO₂ (s) + 2Cl₂ (g)

Therefore, for the theoretical yield, we have;

1 mole of SiCl₄ combines with 1 mole of O₂ to yield 1 mole of SiO₂ and 2 moles of Cl₂

The molar mass of SiCl₄ = 169.9 g/mol

The molar mass of O₂ ≈ 32 g/mol

The molar mass of SiO₂ (s) = 60.08 g/mol

The number of moles of SiCl₄ present in 150 g of SiCl₄ = 150 g/(169.9 g/mol) ≈ 0.883 moles

The number of moles of O₂ present in 200 g of O₂ = 200 g/(32 g/mol) ≈ 6.25 moles

The number of moles of SiO₂ present in 49.2 g of SiCl₄ = 49.2 g/(60.08 g/mol) ≈ 0.818908 moles

Therefore, the theoretical yield of SiO₂ from 0.883 moles of SiCl₄ is 0.883 moles of SiO₂

The mass of 0.833 moles of SiO₂ = 0.833 moles × 60.08 g/mol = 50.04664 grams

The percentage yield = Actual yield/(Theoretical yield) × 100

∴ The percentage yield of SiO₂ for Trial 1 = 49.2 g/(50.04664 g) × 100 ≈ 93.31%

The percentage yield of SiO₂ for Trial 1 ≈ 93.31%

The left over reactant are given as follows;

The number of moles of SiO₂ produced = 0.818908 moles = The number of moles of SiCl₄ used

∴ The number of moles of SiCl₄ remaining = (The number of moles of SiCl₄ present) - (The number of moles of SiCl₄ used)

The number of moles of SiCl₄ remaining = 0.883 moles - 0.818908 moles = 0.064092 moles

The mass of the SiCl₄ remaining = 0.064092 moles × 169.9 g/mol ≈ 10.8892 g

The mass of the SiCl₄ remaining ≈ 10.8892 g

The number of moles of O₂ remaining = (The number of moles of O₂ present) - (The number of moles of O₂ used)

The number of moles of O₂ used in the reaction = The number of moles of SiO₂ produced = 0.818908 moles

∴ The number of moles of O₂ remaining = 6.25 moles - 0.818908 moles = 5.431092 moles

The mass of O₂ remaining = 32 g/mol × 5.431092 moles = 173.794944 grams

The leftover reactants are 10.8892 g of SiCl₄ and 173.794944 g of O₂

b. For trial 2, we have;

The number of moles of SiCl₄ present in 75 g of SiCl₄ = 75 g/(169.9 g/mol) ≈ 0.441436139 moles

The number of moles of O₂ present in 50 g of O₂ = 50 g/(32 g/mol) ≈ 1.5625 moles

The number of moles of SiO₂ present in 25.2 g of SiCl₄ = 25.2 g/(60.08 g/mol) ≈ 0.419440746 moles

Therefore, the theoretical yield of SiO₂ from 0.441436139 moles of SiCl₄ is 0.441436139 moles of SiO₂

The mass of 0.441436139 moles of SiO₂ = 0.441436139 moles × 60.08 g/mol = 26.5214832 grams

The percentage yield of SiO₂ for Trial 2 = 25.2 g/(26.5214832 g) × 100 ≈ 95.02%

Therefore, the ratio of reactants in Trial 2 which is of the ratio SiCl₄ to O₂ of 3:2 is more efficient than that of 3:4 of Trial 1.

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

i belive it is water