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%

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )