Part A Determine the balanced chemical equation for this reaction. C8H18 ( g ) +

Question

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?

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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.

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

.1896octane

0.1896g

0.196g

Explanation / Answer

The balanced reaction is :

2C8H18(g)+25O2(g)16CO2(g)+18H2O(g)

If you have 0.240mole of C8H18(g) and 0.700 mole of O2(g) the limiting reagent is O2(g).

In order to calculate the moles of C8H18(g) in the reaction we have the following:

moles of C8H18(g) in reaction = (2/25) moles of  O2(g) ..

moles of C8H18(g) in reaction = (2/25) (0.700moles) = 0.056 moles of C8H18(g) .

Finally:

moles of C8H18(g) left = moles of C8H18(g) initial - moles of C8H18(g) reaction

moles of C8H18(g) left = 0.240 moles - 0.056moles = 0.184 moles of C8H18(g) left.

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