Liquid octane CH3CH26CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 2.80g of carbon dioxide is produced from the reaction of 3.4g of octane and 5.9g of oxygen gas, calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.


Answer 1




The balanced equation is:

2 CH_(3)(CH_(2))_(6)CH_(3) + 25 O_(2) = 16 CO_(2) + 18 H_(2)O

The first step is to determine the limiting reactant. For this, we calculate the moles of each given component and divide the result for the stoichiometric coefficient.

3.4 g octane / 114.23 g/mol = 0.030 mol octane

0.030 mol octane/2=0.015

5.9 g O2 / 32 g/mol = 0.18 mol O2

0.18 mol O2/25= 0.0074 mol

The lower number, in this case oxygen, is the limiting reactant. The value corresponds to the theoretical yield of the reaction.

Similarly, the real yield is calculated from the product.

2.80 g CO2/ 44.01 g/mol = 0.0636 mol CO2

0.0636 mol CO2/16 = 0.00398 mol

The percent yield is the ratio of the 2 multiplied by a hundred, then

Percent yield= 0.0398/0.0074 *100 = 54%

Answer 2

Final answer:

The percent yield of carbon dioxide from the combustion reaction of octane and oxygen, given the provided masses of the reactants and the yield of CO2, is calculated to be 26.7%


The combustion of octane in oxygen yields carbon dioxide and water in a 1:1 ratio, as the balanced chemical equation for the reaction is 2C8H18 + 25O2 -> 16CO2 + 18H2O. To calculate the percent yield of CO2, we first need to determine the theoretical yield of CO2. We can use the provided masses of octane and O2 and their respective molar masses to calculate the number of moles of each reactant:

- Moles of octane = 3.4 g / 114.22 g/mol = 0.0298 mol
- Moles of O2 = 5.9 g / 32.00 g/mol = 0.184 mol

Now, using the stoichiometric relationship from the balanced chemical equation, we can calculate the theoretical yield of CO2:

- Theoretical yield = 0.0298 mol octane x 16 mol CO2/2 mol octane = 0.238 mol CO2

Next, we convert this to grams using the molar mass of carbon dioxide:

- Theoretical yield = 0.238 mol CO2 x 44.01 g/mol = 10.5g CO2

Now that we have both the actual yield (2.80 g) and the theoretical yield (10.5 g), we can calculate the percent yield:

- Percent yield = (actual yield / theoretical yield) x 100% = (2.80 g / 10.5 g) x 100% = 26.7%

Learn more about Percent Yield here:


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In a 66.0-g aqueous solution of methanol, CH4O, the mole fraction of methanol is 0.290. What is the mass of each component? g



23.84g CH30H

32.81g H2O


We will be using the definition of mole fraction to determine the relationship between the number of moles of methanol,

CH3OH , and the number of moles of water.

But mole fraction gives the ratio between the number of moles of a component i of a solution to the total number of moles present in that solution.


Final answer:

In a 66.0g aqueous solution of methanol with a mole fraction of 0.290, the mass of the methanol is approximately 19.14g and the mass of the water is approximately 46.86g.


In this aqueous solution of methanol (CH4O), we know that its mass is 66.0g and the mole fraction of methanol is 0.290. The mole fraction is defined as the ratio of the number of moles of a component to the total number of moles of all components in the solution.

In order to find the mass of each component, namely the methanol and the water, we first need to establish that if the mole fraction of methanol is 0.290, the mole fraction of water must be 0.710 (because the total of all mole fractions in a solution is always equal to 1).

We then can set up the following proportion: mass of methanol/mass of water = mole fraction of methanol/mole fraction of water. After solving this equation, the mass of methanol will be approximately 19.14g and the mass of the water will be approximately 46.86g.

Learn more about Methanol Aqueous Solution here:


Calculate the pH of a titration at the point when 15.0 mL of 0.15 M NaOH is added to 30.0 mL 0f 0.20 M HNO.



The correct answer is 1.10.


Based on the given information, the molarity of the NaOH is 0.15 M, that is, 0.15 moles per liter of the solution.

Now the moles present in the 15 ml of the solution will be,  

0.015 × 0.15 = 2.25 × 10⁻³ moles of NaOH or 0.0025 moles of NaOH

Now, molarity of the HNO₃ given is 0.20 M, which means 0.2 moles per liter of the solution.  

Now the moles present in the 30 ml of the solution will be,  

0.030 × 0.2 = 0.006 moles of HNO₃

Now the complete disintegration of acid and base will be,  

NaOH (aq) (0.025 moles) ⇔ Na⁺ (aq) (0.025) + OH⁻ (aq) (0.025 moles)

HNO₃ (aq) (0.006 moles) ⇔ H⁺ (0.006 moles) + NO₃⁻ (aq) (0.006 moles)

Now the additional Hydrogen ions at titration point is,  

= 0.006 - 0.0025 = 0.0035 moles of H+

Now the concentration of H+ ions in the 45 ml of the solution will be,  

= 0.0035/45 × 1000

= 0.078 M

pH = -log[H⁺] = -log [0.078]

= 1.10

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the combination of moisture, oxygen and salt ,especially sodium chloride, damages metal worse than rust does. This combination corrodes, or eats away at, the metal, weakening it and causing it to fall apart.

What are the starting substances (molecules) in a chemical equation called?



A chemical reaction is the process in which atoms present in the starting substances rearrange to give new chemical combinations present in the substances formed by the reaction. These starting substances of a chemical reaction are called the reactants, and the new substances that result are called the products.

What type of matter is pepperoni pizza



Heterogeneous Mixture. Have a good day! =)


PLEASE HELP FAST!!If metal X is lower than metal Y on the activity series, then:
A. X will react in water, but only if the temperature is low enough
B. Y will form oxides of X, but only indirectly
C. X will replace ions of Y in a solution
D. Y will replace ions of X in a solution



D. Y will replace ions of X in a solution


If metal X is lower than metal Y on the activity series, then Y will replace ions of X in a solution.

This is the crux of single displacement reactions.

  • In a single displacement reaction, a metallic ion in solution is replaced by a metal higher in the activity series than the metal in solution.
  • On the activity series, metals higher are more reactive and will displace the lower and less reactive ones.
  • Reactivity increases up the group.