How many grams of oxygen are required to react with 10.0 grams of octane (C8H18) in the combustion of octane in gasoline?
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2C8H18+25O2----> 16CO2+18H2O when balanced. 10/114.229 g/mol = .087543... mol, converting to moles by dividing by molar mass. The ratio is 25/2, using the integers in front of the molecules in the balanced equation. so you just multiply .087543... by the ratio. That's 1.09429, which you multiply by the molar mass of O2, 31.9988 g/mol. The answer is 35.02 g O2
35.2 grams of oxygen are required to react with 10.0 grams of octane
Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.
If O₂ is insufficient there will be incomplete combustion produced by CO and H and O
Hydrocarbon combustion reactions (specifically alkanes)
For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:
Balancing C atoms, H and last atoms O atoms
Octane combustion reaction
C₈H₁₈ + O₂ ---> CO₂ + H₂O
To equalize the reaction equation we give the reaction coefficients with variables a, b and c while the most complex compounds, namely Octane, we give the number 1
So the reaction becomes
C₈H₁₈ + aO₂ ---> bCO₂ + cH₂O
C atom on the left 8, right b, so b = 8
left H atom = 18, right 2c so 2c = 18 ---> c = 9
Atom O on the left 2a, right 2b + 2c, so 2a = 2b + 2c
2a = 2.8 + 9
2a = 16 + 9
2a = 25
a = 25/2
The equation becomes:
C₈H₁₈ + 25/2O₂ ---> 8CO₂ + 9H₂O or be
2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O
To find the mass O2, we find the mole first from the mole ratio with Octane
- 1. We are looking for the octane mole
Mr. octane = 8.Ar C + 18.Ar H
Mr. octane = 8.12 + 18.1
Mr octane = 96 + 18
Mr octane = 114
known octane mass: 10 grams then the mole:
mol = gram / Mr
mol = 10/ 114
mol = 0.088
- 2. We look for the O₂ mole
because of the ratio of the reaction coefficient between O2 and octane = 25: 2 then the mole of O₂ =
25/2 x 0.088 = 1,1
So that the mass O₂ =
mole. Mr = 1.1 32
mass = 35.2 grams
a combustion reaction
the type of chemical reaction
The combustion of alkanes, alkenes, and alkynes
Complete combustion of a sample of a hydrocarbon
Keywords: Complete combustion of Hydrocarbons, alkanes, equalize the reaction equation
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B. a decrease in greenhouse gases
C. average decrease in the temperatures near Earth's surface
D. a decrease in carbon dioxide emissions
Answers
Answer: The correct answer is A. average increase in the temperatures near Earth's surface
Explanation: The reason this is correct is that global warming is it getting hotter, logically the temperatures near Earth's surface will be getting hotter as well. Hope this helps
The temperature of the balloon will decrease.
The gas atoms will strike the sides of the balloon more often.
The gas inside the balloon will change into a liquid.
Answers
Answer: The gas atoms will strike the sides of the balloon more often.
Explanation:
Kinetic energy is the energy possessed by an object by virtue of its motion.
Average kinetic energy is defined as the average of the kinetic energies of all the particles present in a system. It is determined by the equation:
K= kinetic energy
R= gas constant
T= temperature
From above, it is visible that kinetic energy is directly related to the temperature of the system.
So if average kinetic energy of the system increases , the speed of the molecules will be more and hence the gas atoms will strike the sides of the balloon more often.
Hope I helped!
Answers
Answers
Answers
Answer:
T₂ = 379.4 K
Explanation:
Given data:
Initial volume = 1.56 L
Initial temperature = 20°C (20+273 = 293 K)
Final volume = 2.02 L
Final temperature = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ /V₁
T₂ = 2.02 L × 293 K / 1.56 L
T₂ = 591.86 L.K / 1.56 L
T₂ = 379.4 K