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Determine the balanced chemical equation for this reaction. C8H18(g)+O2(g)→CO2(g)+H2O(g)

Part A:
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

Part B:
0.280 mol of octane is allowed to react with 0.630 mol of oxygen. Which is the limiting reactant?

Oxygen

Part C:
How many moles of water are produced in this reaction?

H2O produced = 0.454 mol
Part D:
After the reaction, how much octane is left?

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Answer: balanced reaction equation

C8H18(g) + 25/2 O2(g) ---------> 8CO2(g) + 9H2O(l)

Explanation:

Part A- coefficients

1, 25/2,8,9

Part B

Oxygen is the limiting reactant

Part C

If 1 mole of octane produced 9 moles of water from the balanced reaction equation

0.28 moles of octane will produce 0.28×9= 2.52 moles of water

Part D

If 12.5 moles of oxygen reacts with 1 mole of octane

0.63 moles of oxygen will react with 0.63/12.5=0.0504moles of octane

Amount of octane left= 0.280-0.0504=0.2296 moles

Part A:

The balanced chemical equation for the reaction is

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

The coefficients for each compound in order is 2, 25, 16, 18

Part B:

The limiting reactant is Oxygen

Part C:

The number of mole of water produced is 0.454 mol

Part D:

The number of moles of octane left is 0.230 mol

  • Part A

For the reaction,

The balanced chemical equation for the reaction is

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

This means 2 moles of C₈H₁₈ will react with 25 moles of O₂ to produce 16 moles of CO₂ and 18 moles of H₂O

Hence, the coefficients for each compound in order is 2, 25, 16, 18

  • Part B

If 0.280 mol of octane is allowed to react with 0.630 mol of oxygen,

To determine the limiting reagent, we will calculate the number of moles of oxygen that is required to react with the 0.280 mol of octane

Since,

2 moles of octane reacts with 25 moles of oxygen,

Then,

0.280 mole of octane will react with [tex]\frac{25 \times 0.280}{2}[/tex] moles of oxygen

[tex]\frac{25 \times 0.280}{2} = 3.5[/tex]

∴ 3.5 moles of oxygen is required to react with the 0.280 mole of octane.

Since the number of moles of oxygen present (0.630 mole) is less than the required amount of oxygen needed (3.5 mole), then, oxygen is the limiting reactant

Hence, the limiting reactant is Oxygen

  • Part C

To determine the number of moles of water produced,

First, we will calculate the number of moles of octane that reacted

From the balanced chemical equation

2 moles of octane reacts with 25 moles of oxygen

Then,

[tex]\frac{0.630 \times 2}{25}[/tex] moles of octane will react with the 0.630 mole of oxygen present

[tex]\frac{0.630 \times 2}{25}= 0.0504[/tex]

∴ Only 0.0504 mole of octane reacted

Now,

If 2 moles of octane react with 25 moles of O₂ to produce 18 moles of water,

Then,

0.0504 mole of octane will react with 0.630 mole of oxygen to produce 0.0504 × 9 mole of water

0.0504 × 9 = 0.4536 mol ≅ 0.454 mol

Hence, the number of mole of water produced is 0.454 mol

  • Part D

For the quantity of octane that is left

Number of moles of octane left = Number of moles of octane present - Number of moles of octane that reacted

Number of moles of octane left = 0.280 mol - 0.0504 mol

Number of moles of octane left = 0.2296 mol

Number of moles of octane left ≅ 0.230 mol

Hence, the number of moles of octane left is 0.230 mol

Learn more here: https://brainly.com/question/16934894

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