# Problem: 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.Part B:0.280 mol of octane is allowed to react with 0.630 mol of oxygen. Which is the limiting reactant?Part C:How many moles of water are produced in this reaction?Part D:After the reaction, how much octane is left?

###### FREE Expert Solution

Step 1: From the problem, the given combustion reaction is:

C8H18 + O2 → CO2 + H2O

This equation is not yet balanced. To balance it, we have to make sure that the number of elements on both sides is equal.

Balance C: We have 8 C on the reactant side and 1 C on the product side – add a coefficient of 8 to CO2:

C8H18 + O2 → 8 CO2 + H2O

Balance H: We have 18 H on the reactant side and 2 H on the product side – add a coefficient of 9 to H2O:

C8H18 + O2 → 8 CO2 +  9 H2O

Balance O: We have 2 O on the reactant side and 8(2) + 9 = 25 O on the product side – multiply the entire equation by 2 and add a coefficient of 25 to O2:

C8H1825 O2 → 16 CO218 H2O

Now that we have a balanced equation, we can proceed with the problem.

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###### Problem Details

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.

Part B:

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

Part C:

How many moles of water are produced in this reaction?

Part D:
After the reaction, how much octane is left?