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hjlf
3 years ago
13

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.
Chemistry
2 answers:
miss Akunina [59]3 years ago
5 0

Answer:

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

B. 761.42 L

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

Lana71 [14]3 years ago
4 0

Answer:

a. C_{12}H_{26}(l)+\frac{37}{2}O_2(g)\rightarrow 12CO_2(g)+13H_2O(g)

b. V=761 L

Explanation:

Hello,

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

C_{12}H_{26}(l)+\frac{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}*\frac{1000gC_{12}H_{26}}{1kgC_{12}H_{26}}*\frac{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}*\frac{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=\frac{nRT}{p}=\frac{31.8mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{1atm}\\\\V=761 L

Best regards.

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How many moles are in 2.5L of 1.75 M Na2CO3
Natasha_Volkova [10]

Answer:

4.4 mol.

Explanation:

Hello!

In this case, since the formula for calculating the molarity is:

M=n/V

Whereas n stands for moles and V for the volume in liters; we can solve for n as shown below when we are given the volume and the molarity:

n=V*M

Thus, we plug in the given data to obtain:

n=2.5L*1.75mol/L=4.4mol

Best regards!

4 0
2 years ago
Read 2 more answers
Which of the following statements is true of an aqueous solution of sodium chloride?
Ilya [14]

Answer:

What are the statements please

Explanation:

4 0
3 years ago
You have a graduated cylinder with 10 mL of water in it.
Law Incorporation [45]

Answer:

<h3>The answer is 10 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 300 g

volume = final volume of water - initial volume of water

volume = 40 - 10 = 30 mL

We have

density =  \frac{300}{30}  =    \frac{30}{3}  \\

We have the final answer as

<h3>10 g/mL</h3>

Hope this helps you

5 0
3 years ago
given two equations representing reactions: which type of reaction is represented by each of these equations?
Sergeeva-Olga [200]

Answer:

Equation 1 - nuclear fission

Equation 2 - nuclear fusion

Explanation:

Nuclear fission is a reaction in which a large nucleus is split into smaller nuclei when it is bombarded by neutrons. The process produces more neutrons to continue the chain reaction. This is clearly depicted in equation 1 as shown in the question.

Nuclear fusion is a reaction in which two light nuclei combine in order to form a larger nuclei. This is clearly depicted in equation 2 as shown in the question.

7 0
2 years ago
At a given temperature the vapor pressures of hexane and octane are 183 mmhg and 59.2 mmhg, respectively. Calculate the total va
Bas_tet [7]

Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

P=X_{A}P_{A}+X_{B}P_{B}

Here, X_{A} is mole fraction of A, X_{B} is mole fraction of B, P_{A} is partial pressure of A and P_{B} is partial pressure of B.

The mole fraction of A and B are related to each other as follows:

X_{A}+X_{B}=1

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

X_{octane}=1-X_{hexane}=1-0.58=0.42

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.

Now, vapor pressure can be calculated as follows:

P=X_{hexane}P_{hexane}+X_{octane}P_{octane}

Putting the values,

P=(0.580)(183 mmHg)+(0.420)(59.2 mmHg)=131 mmHg

Therefore, total vapor pressure over the solution of hexane and octane is 131 mmHg.

4 0
3 years ago
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