The reaction between 32.0 g of CH4 and excess oxygen gas produces 75.9 g of CO2 gas and some water. Determine the percent yield.

The percent yield of the reaction is 86.2 %

Explanation:

Step 1: Data given

Mass of CH4 = 32.0 grams

Molar mass CH4 = 16.04 g/mol

Mass of CO2 = 75.9 grams

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

CH4 + 2O2 → CO2 + 2H2O

Step 3: Claculate moles CH4

Moles CH4 = mass CH4 / molar mass CH4

Moles CH4 = 32.0 grams / 16.04 g/mol

Moles CH4 = 2.00 moles

Step 4: Calculate moles CO2

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

For 2.00 moles CH4 we'll have 2.00 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 2.00 moles * 44.01 g/mol

Mass CO2 = 88.02 grams

Step 6: Calculate percent yield

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

Percent yield = (75.9 grams / 88.02 grams) * 100%

PErcent yield = 86.2 %

The percent yield of the reaction is 86.2 %

86.25%

Explanation:

Step 1:

The balanced equation for the reaction between CH4 and O2. This is given below:

CH4 + 2O2 —> CO2 + 2H2O

Step 2:

Determination of the mass of CH4 that reacted and the mass of CO2 produced from the balanced equation. This is shown below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

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

From the balanced equation:

16g of CH4 reacted.

44g of CO2 is produced.

Step 3:

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

From the balanced equation above,

16g of CH4 produced 44g of CO2.

Therefore, 32g of CH4 will produce = (32x44)/16 = 88g of CO2.

Therefore, the theoretical yield of CO2 is 88g.

Step 4:

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

Actual yield = 75.9g

Theoretical yield = 88g

Percentage yield =?

Percentage yield = Actual yield /Theoretical yield x100

Percentage yield = 75.9/88 x 100

Percentage yield = 86.25%

Therefore, the percentage yield of CO2 is 86.25%

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

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