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ira [324]
3 years ago
14

Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.

If 51.9 g of carbon dioxide is produced from the reaction of 34.29 g of butane and 165.7 g of oxygen gas, calculate of the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it. (first balance the equation below) CH3(CH2)2CH3+O2−−−>CO2+H2O
Chemistry
1 answer:
weeeeeb [17]3 years ago
7 0

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

OR

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

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∵ no. of moles of Fe = mass of Fe/atomic weight of Fe.

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If 42 grams of carbon and 52 grams of oxygen are used, how many grams of CO2 will be produced (Hint: find the
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Answer:

71.5g

Explanation:

The reaction equation is given as:

               C  +  O₂  →  CO₂

Mass of C = 42g

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Unknown:

Mass of CO₂ produced  = ?

Solution

Now to solve this problem, we have to find limiting reactant which is the one given in short supply in this reaction.

 The extent of the reaction is controlled by this reactant.

Find the number of moles of the given species;

 Number of moles  = \frac{mass}{molar mass}

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Now;

   From the balanced reaction equation;

           1 mole of C reacted with 1 mole of O₂

We see that C is in excess and O₂ is the limiting reactant.

            1 mole of O₂ will produce 1 mole of CO₂

 So;      1.63mole of O₂ will produce 1.63 mole of CO₂

Mass of CO₂ = number of moles x molar mass

       Molar mass of CO₂ = 44g/mol

Mass of CO₂ = 1.63 x 44 = 71.5g

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