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?

	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

Answer 1

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