Question

Suppose 1.65 moles of C₆H₆ react with excess oxygen to produce carbon dioxide and water.

Answer 1

Answer:

1 9.9

2 4.95

3 12.375

Explanation:

To solve this problem, we basically need to write a balanced chemical equation:

Like all other hydrocarbons, benzene burns in oxygen to yield water and carbon iv oxide as the only products. A balanced chemical equation is given below:

2C6H6(l)+15O2(g)→12CO2(g)+6H2O(g)

Let’s now proceed to answer the questions.

A. Number of moles of carbon iv oxide produced

From the balanced chemical equation, we can see that one mole of benzene yielded 6 moles of carbon iv oxide.

This means 1.65 moles of benzene will yield: 6 * 1.65 = 9.9 moles of CO2

B. From the balanced equation, we can see that one mole of benzene yielded 3 moles of water. Hence, 1.65 moles will yield 1.65 * 3 = 4.95 moles of water

C. To find the number of moles of oxygen consumed, we can see that 2 moles of benzene consumed 15 moles of oxygen. 1.65 moles of benzene would thus consume 1.65 * 15/2moles = 12.375moles of oxygen

Answer 2

Answer:

A. 9.9 moles of CO2.

B. 4.95 moles of H2O.

C. 12.375 moles of O2

Explanation:

Step 1:

The equation for the reaction is given below:

C6H6 + O2 —> CO2 + H2O

Step 2:

Balancing the equation. This is illustrated below :

C6H6 + O2 —> CO2 + H2O

There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 6 in front of CO2 as shown below:

C6H6 + O2 —> 6CO2 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

C6H6 + O2 —> 6CO2 + 3H2O

There are a total of 15 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 15/2 in front of O2 as shown below:

C6H6 + 15/2O2 —> 6CO2 + 3H2O

Multiply through by 2 to clear the fraction as shown below:

2C6H6 + 15O2 —> 12CO2 + 6H2O

Now the equation is balanced.

A. Step 3:

Determination of the moles of CO2 produced when 1.65 moles of C6H6 react with excess oxygen This is illustrated below:

2C6H6 + 15O2 —> 12CO2 + 6H2O

From the balanced equation above,

2 moles of C6H6 produced 12 moles of CO2.

Therefore, 1.65 moles of C6H6 will produce = (1.65x12)/2 = 9.9 moles of CO2.

B. Step 4:

Determination of the moles of H2O produced when 1.65 moles of C6H6 react with excess oxygen This is illustrated below:

2C6H6 + 15O2 —> 12CO2 + 6H2O

From the balanced equation above,

2 moles of C6H6 produced 6 moles of H2O.

Therefore, 1.65 moles of C6H6 will produce = (1.65x6)/2 = 4.95 moles of H2O.

C. Step 5:

Determination of the moles of O2 consumed during the reaction. This is illustrated below:

2C6H6 + 15O2 —> 12CO2 + 6H2O

From the balanced equation above,

2 moles of C6H6 consumed 15 moles of O2.

Therefore, 1.65 moles of C6H6 will consume = (1.65x15)/2 = 12.375 moles of O2