# Question 3870a

Jun 26, 2015

That much propane would form 130 kg of carbon dioxide.

#### Explanation:

Start by writing the balanced chemcial equation for the combustion of propane, ${C}_{3} {H}_{8}$

${C}_{3} {H}_{8 \left(g\right)} + 5 {O}_{2 \left(g\right)} \to \textcolor{red}{3} C {O}_{2 \left(g\right)} + 4 {H}_{2} {O}_{\left(g\right)}$

Notice that you have a $1 : \textcolor{red}{3}$ mole ratio between propane and carbon dioxide. This means that, regardless of how many moles of propane react, the reaction will always produce 3 times more moles of carbon dioxide.

Assuming that the oxygen is in excess, you can determine how many moles of propane react by using propane's molar mass

42cancel("kg") * (1000cancel("g"))/(1cancel("kg")) * "1 mole"/(44.1cancel("g")) = "952.4 moles" ${C}_{3} {H}_{8}$

This means that the reaction will produce

952.4cancel("moles"C_3H_8) * (color(red)(3)" moles "CO_2)/(1cancel("mole"C_3H_8)) = "2857.2 moles" $C {O}_{2}$

To determine the mass of the produced carbon dioxide, use carbon dioxide's molar mass

2857.2cancel("moles") * "44.01 g"/(1cancel("mole")) = "125,745.4 g"# $C {O}_{2}$

Rounded to two sig figs, the number of sig figs you gave for the mass of propane, and expressed in kilograms, the answer will be

$125 , 754.4 \cancel{\text{g") * "1 kg"/(1000cancel("g")) = color(green)("130 kg}}$