Part A
Determine the balanced chemical equation for this reaction.
C8H18(g)+O2(g)?CO2(g)+H2O(g)
Enter the coefficients for each compound in order, separated by commas. For example, 1,2,3,4 would indicate one mole of C8H18, two moles of O2, three moles of CO2, and four moles of H2O.
2,25,16,18
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It is important to balance a chemical equation before using it for calculations. Checking that equations are balanced will help you avoid many errors in chemistry problems.
Balanced chemical equation
2C8H18(g)+25O2(g)?16CO2(g)+18H2O(g)
Part B
0.240 mol of octane is allowed to react with 0.700 mol of oxygen. Which is the limiting reactant?
octane | |
oxygen |
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Now that you have identified oxygen as the limiting reactant, you can use the number of moles of oxygen to find the numbers of moles of all the other substances.
Part C
How many moles of water are produced in this reaction?
Express your answer with the appropriate units.
H2O produced = |
0.504 mol |
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Part D I need the anwser for part D
After the reaction, how much octane is left?
these all of my wrong answers
= |
0.777mole |
moles of C8H18 remaining = |
.1896octane |
moles of C8H18 remaining = |
0.1896g |
moles of C8H18 remaining = |
0.196g |
Part B
Given
0.240 mol of octane is allowed to react with 0.700 mol of oxygen.
0.24 mole octane * 25 mole O 2 / 2 mole octane = 3 mole O2
And
0.700 mole O2 *2 mole Octane / 25 mole O2= 0.056 mole Octane
Oxygen is a limiting agent while octane present in excess
The limiting reagent (or limiting reactant) is a substance that limits the amount of product in the reaction. This limiting agent is completely consumed in the reaction.
Part C
0.700 mole O2 * 18 mole H2O / 25 mole O2= 0.504 moel H2O
Part D:
Moles of octane which is used:
0.700 mole O2 *2 mole Octane / 25 mole O2= 0.056 mole Octane
Left mole of octane = 0.240 – 0.056 = 0.184 mole
Amount in g = number of mole s* molar mass
= 0.184 * 114.23 g/ molle
= 21.01832 g
= 21.0 g
Part A Determine the balanced chemical equation for this reaction. C8H18(g)+O2(g)?CO2(g)+H2O(g) Enter the coefficients for each...
background info: So I have the balanced equation = 2C8H18(g)+25O2(g)→16CO2(g)+18H2O(g) and oxygen is the limiting reactant of 0.180 mol of octane is allowed to react with 0.880 mol of oxygen. Part C How many moles of water are produced in this reaction? Express your answer with the appropriate units. View Available Hint(s) : HÅR O 2 ? H2O produced = H,0 produced = Value Units Submit Part D After the reaction, how much octane is left? Express your answer with...
Balanced chemical equation: 2C8H18(g)+25O2(g)->16CO2(g)+18H2O(g) 0.320 mol of octane is allowed to react with 0.880 mol of oxygen. How many moles of water are produced in this reaction?
balanced chemical equation: 2C8H18(g)+25O2->16CO2(g)+18H2O(g) 0.320 mol of octane is allowed to react with 0.880 mol of oxygen. After the reaction how much Octane is left?
The combustion of octane, C8H18,C8H18, proceeds according to the reaction shown. 2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l) If 522 mol522 mol of octane combusts, what volume of carbon dioxide is produced at 38.0 ∘C38.0 ∘C and 0.995 atm?
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Consider the combustion reaction for octane (C8H18), which is a primary component of gasoline. 2C8H18+25O2⟶16CO2+18H2O How many moles of CO2 are emitted into the atmosphere when 26.6 g C8H18 is burned?
Consider the combustion reaction for octane (C8H18), which is a primary component of gasoline. 2C8H18+25O2⟶16CO2+18H2O How many moles of CO2 are emitted into the atmosphere when 25.6 g C8H18 is burned?
The complete combustion of octane, C8H18, a component of gasoline, proceeds as follows: 2C8H18(l)+25O2(g)→16CO2(g)+18H2O(g) a. How many moles of O2 are needed to burn 1.35 mol of C8H18? b. How many grams of O2 are needed to burn 13.0 g of C8H18? c. Octane has a density of 0.692 g/mL at 20 degrees C. How many grams of O2 are required to burn 1.20 gal of C8H18?
Let’s combust octane. (2C8H18 + 25O2 --> 16CO2 + 18H2O) delta G C8H18= 16.4 kj/mol, delta G CO2 = -394 kj/mol, delta G H2O = -229 kj/mol a. Please calculate the free energy of rxn for the burning of octane. b. Now give the lnKeq at STP. c. Next, give the lnKeq at 200C d. Follow with the free energy of reaction at this temperature
The combustion of octane, C8H18, proceeds according to the reaction shown. 2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l) Calculate the volume of carbon dioxide produced when 562 moles of octane combusts at 28.0 °C and 1.00 bar.