Propane, C_3H_8, reacts with oxygen to produce carbon dioxide and water. If 10.0 grams of oxygen reacts with an excess of propane, then how many moles of carbon dioxide were produced?

Jan 25, 2016

The reaction will form $\text{0.188 mol}$ of ${\text{CO}}_{2}$.

Explanation:

Given: Mass of ${\text{O}}_{2}$; chemical equation (understood)

Find: Moles of ${\text{CO}}_{2}$

Strategy:

The central part of any stoichiometry problem is to convert moles of something to moles of something else.

(b) We can use the molar ratio from the equation to convert moles of ${\text{O}}_{2}$ to moles of ${\text{CO}}_{2}$.

${\text{moles of O"_2stackrelcolor (blue)("molar ratio"color(white)(Xl))( →) "moles of CO}}_{2}$

(c) Then we can use the molar mass to convert the mass of ${\text{O}}_{2}$ to moles of ${\text{CO}}_{2}$.

Our complete strategy is:

${\text{Mass of O"_2stackrelcolor (blue)("molar mass"color(white)(Xl))(→) "moles of O"_2stackrelcolor (blue)("molar ratio"color(white)(Xl))( →) "moles of CO}}_{2}$

Solution

(a) The balanced equation is

$\text{C"_3"H"_8 + "5O"_2 → "3CO"_2 + "4H"_2"O}$

(b) Calculate moles of ${\text{O}}_{2}$

10.0 color(red)(cancel(color(black)("g O"_2))) × ("1 mol O"_2)/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.3125 mol O"_2

(c) Calculate moles of ${\text{CO}}_{2}$

The molar ratio of ${\text{CO}}_{2}$ to ${\text{O}}_{2}$ is ("3 mol CO"_2)/("5 mol O"_2)"

${\text{Moles of CO"_2 = 0.3125 color(red)(cancel(color(black)("mol O"_2))) × ("3 mol CO"_2)/(5 color(red)(cancel(color(black)("mol O"_2)))) = "0.188 mol CO}}_{2}$
(3 significant figures)

Answer: The reaction will form $\text{0.188 mol}$ of ${\text{CO}}_{2}$.