Consider the following balanced equation for the following reaction:
15O2(g) + 2C6H5COOH(aq) → 14CO2(g) + 6H2O(l)
Determine the amount of CO2(g) formed in the reaction if the percent yield of CO2(g) is 83.0% and the theoretical yield of CO2(g) is 1.30 moles.

47.47 g of CO₂ is the amount formed.

Explanation:

The reaction is:

2C₆H₅COOH(aq) + 15O₂(g) → 14CO₂(g) + 6H₂O(l)

Let's apply the formula for the percent yield

Percent yield of reaction = (Produced yield/Theoretical yield) . 100

First of all we convert the moles of CO₂ to mass: 1.30 mol . 44 g /1 mol = 57.2 g. So now, we replace:

(Produced yield / 57.2 g ). 100 = 83

Produced yield / 57.2 g  = 83 / 100

Produced yield / 57.2 g  = 0.83

Produced yield = 0.83 . 57.2g → 47.47 g of CO₂

The actual yield of CO2 is 1.079 moles or 47.5 grams formed

Explanation:

Step 1: Data given

Number of CO2 = 1.30 moles

Percent yield = 83.0 %

Step 2: The balanced equation

15O2(g) + 2C6H5COOH(aq) → 14CO2(g) + 6H2O(l)

Step 3: Calculate the number moles of CO2 formed

1.30 moles CO2 = 100 %

The actual amount of moles = 0.83 * 1.30 = 1.079 moles

Step 4: Calculate the percent yield of the reaction

We can control this by calculating the percent yield of the reaction

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

% yield = (1.079 moles / 1.30 moles ) * 100 %

% yield = 83.0 %

Step 5: Calculate thr mass of CO2 produced

Mass CO2 = moles * molar mass CO2

Mass CO2 = 1.079 moles * 44.01 g/mol

Mass CO2 = 47.5 grams

The actual yield of CO2 is 1.079 moles or 47.5 grams formed

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.