Combustion of Octane:
C8H18 + O2 → CO2 + H2O
Question 1: What are the coefficients for the balanced reaction of the combustion of octane?
Problem 2: If 20 g of octane combust with 20 g of oxygen, which is the limiting reagent?
Q1)
aC8H18 + bO2 --> cCO2 + dH2O
Assuming the stoichiometric ratios in the reaction to be a, b,
c, and d, we have:
8a = c
18a = 2d
=> 9a = d
2b = 2c + d
Therefore,
2b = 2(8a) + 9a
=> 2b = 16a + 9a = 25a
=> b = 12.5 a
Therefore,
b = 12.5 a
c = 8a
d = 9a
The reaction is:
a C8H18 + 12.5a O2 --> 8a CO2 + 9a H2O
Multiplying by 2;
2a C8H18 + 25a O2 -->
16a CO2 + 18a H2O
or:
2 C8H18 + 25 O2 -->
16 CO2 + 18 H2O
Q2)
If there are 20g of Octane and 20g of Oxygen is present;
No. of moles of Octane = Mass/Mol. Mass = 20/114.232 = 0.175
mol
No of moles of Oxygen = Mass/Mol. Mass = 20/32 = 0.625 mol
Now, from the equation in Q1, for 2 mol of Octane, 25 mol of Oxygen is required for complete combustion. Since 20 grams of Octane is 0.175 mol, we would require 2.1875 mol of Oxygen for complete combustion. However, only 0.625 mol of Oxygen is present. Therefore, Oxygen is the limiting reagent.
Octane (C8H18) reacts with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). When the equation in balanced, the coefficient of octane is: C8H18+ O2 --> CO2 + H2O a.8 b. 16 c.25 d. 2
The combustion of octane (C8H18) in oxygen proceeds as follows 2 C8H18(g) + 25 O2(g) ---> 16 CO2 (g) + 18 H2O(l) How many moles of CO2are produced when 5.0 moles of octane, C8H18, is burned in 5.0 moles of oxygen?
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 SubmitHintsMy AnswersGive UpReview Part Correct 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...
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