Question

Give the percent yield when 162.8 g of CO2 are formed from the reaction of excess amount of

Answer 1

Answer:

84.86%

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction between C8H18 and O2 to produce CO2. This is illustrated below:

2C8H18 + 25O2 —> 16CO2 + 18H2O

Step 2:

Now, let us calculate the mass of O2 that reacted and the mass of CO2 produced from the balanced equation above. This is illustrated below:

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 25 x 32 = 800g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 16 x 44 = 704g

Therefore the mass of O2 that reacted from the balanced equation is 800g

The mass of CO2 produced from the balanced equation is 704g

Step 3:

Determination of the theoretical yield of CO2. This is illustrated below:

From the balanced equation above,

800g of O2 reacted to produced 704g of CO2.

Therefore, 218g of O2 will react to produce = (218 x 704)/800 = 191.84g of CO2.

Therefore, the theoretical yield of CO2 is 191.84g

Step 4:

Determination of the percentage yield of CO2. This is illustrated below:

Actual yield = 162.8g

Theoretical yield = 191.84g

Percentage yield =?

Percentage yield = Actual yieldm/Theoretical yield x100

Percentage yield = 162.8/191.84 x100

Percentage yield = 84.86%

Therefore, the percentage yield of CO2 is 84.86%

Answer 2

Answer:

The percent yield of the reactions is  84.8 %

Explanation:

Step 1: data given

Mass of CO2 formed = 162.8 grams

Mass of O2 = 218.0 grams

C8H18 is in excess

Molar mass O2 = 32.0 g/mol

Molar mass of CO2 = 44.01 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles O2

Moles O2 = 218.0 grams / 32.0 g/mol

Moles O2 = 6.8125 moles

Step 4: Calculate moles CO2

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

For 6.8125 moles O2 we'll have 16/25 * 6.8125 = 4.36 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 4.36 moles ¨¨ 44.01 g/mol

Mass CO2 = 191.9 grams

Step 6: Calculate the percent yield

Percent yield = (actual mass / theoretical mass) * 100 %

Percent yield = (162.8 grams / 191.9 grams ) * 100 %

Percent yield = 84.8 %

The percent yield of the reactions is  84.8 %