Suppose 1.65 moles of c₆h₆ react with excess oxygen to produce carbon dioxide and water.
a. how many moles of carbon dioxide will be produced in this reaction?
b. how many moles of water will be produced in this reaction?
c. how many moles of oxygen gas will be consumed during the reaction?

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

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

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

2b2t

explanation:

2b2t