Question

Combustion of hydrocarbons such as dodecane (C12H26) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.

a. Write a balanced chemical equation, including physical state symbols, for the combustion of liquid dodecane into gaseous carbon dioxide and gaseous water.


b. Suppose 0.450 kg of dodecane are burned in air at a pressure of exactly 1 atm and a temperature of 19.0 O. Calculate the volume of carbon dioxide gas that is produced. Be sure your answer has the correct number of significant digits.

Answer 1

a.
`C_(12)H_(26)(l)+(37)/(2)O_2(g)\rightarrow 12CO_2(g)+13H_2O(g)`

b.
`V=761 L`

Step-by-step explanation:

Hello,

a. In this case, the required balanced combustion is:

`C_(12)H_(26)(l)+(37)/(2)O_2(g)\rightarrow 12CO_2(g)+13H_2O(g)`

b. In this case, we compute the available moles of dodecane as:

`n_{C_(12)H_(26)}^(available)=0.450kgC_(12)H_(26)*(1000gC_(12)H_(26))/(1kgC_(12)H_(26))*(1molC_(12)H_(26))/(170gC_(12)H_(26)) =2.65molC_(12)H_(26)`

Next, we compute the yielded moles of carbon dioxide by using the 1:12 mole ratio given at the chemical reaction:

`n_(CO_2)=2.65molC_(12)H_(26)*(12molCO_2)/(1molC_(12)H_(26)) =31.8molCO_2`

Finally, by using the ideal gas equation we compute the volume at the given conditions:

`V=(nRT)/(p)=(31.8mol*0.082(atm*L)/(mol*K)*(19+273.15)K)/(1atm)\\\\V=761 L`

Best regards.

Answer 2

A. 2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

B. 761.42 L

Step-by-step explanation:

A. Step 1:

The equation for the reaction.

C12H26(l) + O2(g) —> CO2(g) + H2O(g)

A. Step 2:

Balancing the equation.

The equation can be balance as follow:

C12H26(l) + O2(g) —> CO2(g) + H2O(g)

There are 12 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 12 in front of CO2 as illustrated below:

C12H26(l) + O2(g) —> 12CO2(g) + H2O(g)

There are 26 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 13 in front of H2O as illustrated below:

C12H26(l) + O2(g) —> 12CO2(g) + 13H2O(g)

Now, there are a total of 37 atoms of O2 on the right side and 2 atoms on the left. It can be balance by putting 37/2 in front of O2 as illustrated below:

C12H26(l) + 37/2O2(g) —> 12CO2(g) + 13H2O(g)

Multiply through by 2 to clear the fraction from the equation.

2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

Now the equation is balanced

B. Step 1:

We'll by obtaining the number of mole of C12H26 in 0.450 kg of C12H26. This is illustrated below:

Molar Mass of C12H26 = (12x12) + (26x1) = 144 + 26 = 170g/mol

Mass of C12H26 = 0.450 kg = 0.450x1000 = 450g

Number of mole of C12H26 =?

Number of mole = Mass/Molar Mass

Number of mole of C12H26 = 450/170

Number of mole of C12H26 = 2.65 moles

B. Step 2:

Determination of the number of mole of CO2 produced by the reaction. This is illustrated below:

2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

From the balanced equation above,

2 moles of C12H26 produced 24 moles of CO2.

Therefore, 2.65 moles of C12H26 will produce = (2.65x24)/2 = 31.8 moles of CO2.

B. Step 3:

Determination of the volume of CO2 produced by the reaction.

Pressure (P) = 1 atm

Temperature (T) = 19°C = 19°C + 273 = 292K

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n) = 31.8 moles

Volume (V) =?

The volume of CO2 produced by the reaction can b obtained by applying the ideal gas equation as follow:

PV = nRT

1 x V = 31.8 x 0.082 x 292

V = 761.42 L

Therefore, the volume of CO2 produced is 761.42 L