How many molecules of O_2 are required for the complete combustion of 23.0 grams of hexane in the reaction C_6H_14 + O_2 -> CO_2 + H_2O?

Mar 26, 2016

The complete combustion requires 1.527× 10^24color(white)(l) "molecules O"_2.

There are four steps to follow in this problem:

1. Balance the equation.
2. Use the molar mass of hexane to calculate the moles of hexane.
3. Use the molar ratio of oxygen:hexane from the balanced equation to calculate the moles of oxygen.
4. Use Avogadro's number to calculate the molecules of oxygen.

Let's get started.

1. Balance the equation

$\text{2C"_6"H"_14 + "19O"_2→ "12CO"_2+"14H"_2"O}$

2. Calculate the moles of hexane

${\text{Moles of C"_6"H"_14 = 23.0 color(red)(cancel(color(black)("g C"_6"H"_14))) × ("1 mol C"_6"H"_14)/(86.17 color(red)(cancel(color(black)("g C"_6"H"_14)))) = "0.267 mol C"_6"H}}_{14}$

3. Calculate the moles of oxygen

${\text{Moles of O"_2 = 0.267 color(red)(cancel(color(black)("mol C"_6"H"_14))) × ("19 mol O"_2)/(2 color(red)(cancel(color(black)("mol C"_6"H"_14)))) = "2.54 mol O}}_{2}$

4. Finally, we calculate the molecules of oxygen

"Molecules of O"_2 = 2.54 color(red)(cancel(color(black)("mol O"_2))) × (6.022 × 10^23color(white)(l) "molecules O"_2)/(1 color(red)(cancel(color(black)("mol O"_2))))
= 1.527 × 10^24color(white)(l) "molecules O"_2