Question

3.9 g of ${\text{C}}{{\text{O}}_{\text{2}}}$ is produced from how many liters of ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}$?

Answer 1

We know that, when a hydrocarbon (methane, propane etc.) undergoes reaction with water, it produces carbon dioxide and water. This reaction is termed a combustion reaction. Here, first we have to write the combustion reaction of propane.

Complete step by step answer:
The combustion reaction of propane is,
${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}} + 5{{\text{O}}_{\text{2}}} \to 3{\text{C}}{{\text{O}}_{\text{2}}} + 4{{\text{H}}_{\text{2}}}{\text{O}}$
From the above combustion reaction of propane, we find that, for one mole of propane we get three moles of carbon dioxide. So, the mole ratio of ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}$ and ${\text{C}}{{\text{O}}_{\text{2}}}$ is 1:3.
Given that, the mass of carbon dioxide produced is 3.9 g. Now, we have to calculate the moles of carbon dioxide produced. The molar mass of carbon dioxide is $44\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$.
Moles of carbon dioxide produced=$\dfrac{{{\text{Mass}}}} {{{\text{Molar}}\,{\text{mass}}}} = \dfrac{{3.9\,{\text{g}}}} {{44\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}} = 0.09\,{\text{mol}}$
As the mole ratio of ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}$ and ${\text{C}}{{\text{O}}_{\text{2}}}$ is 1:3 is, so, the moles of ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}$ (propane) required to produce 0.09 moles of carbon dioxide is $\dfrac{{0.09}}{3} = 0.03$mol
Now, we have to convert the number of moles of propane to grams of propane. For that, we have to use the formula,
Number of moles=$\dfrac{{{\text{Mass}}}} {{{\text{Molar}}\,{\text{mass}}}}$
$\Rightarrow 0.0.3\,\,{\text{mol}} = \dfrac{{{\text{Mass}}}} {{44.1\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}$
$\Rightarrow {\text{Mass}} = {\text{1}}{\text{.32}}\,{\text{g}}$
So, the mass of propane needed to produce 3.9 g of carbon dioxide is 1.32 g.
Now, we have to calculate the volume of the propane. For that, we have to use the density.
Density=$\dfrac{{{\text{Mass}}}} {{{\text{Volume}}}}$
The mass of propane is 1.32 g and the density of propane is $2.01\,{\text{kg/}}{{\text{m}}^{\text{3}}} = 0.00201\,{\text{g/c}}{{\text{m}}^{\text{3}}}$
. We have to put these values in the formula of density.
$\Rightarrow 0.00201{\text{g/c}}{{\text{m}}^{\text{3}}} = \dfrac{{1.32\,{\text{g}}}} {{{\text{Volume}}}}$
$\Rightarrow {\text{Volume}} = {\text{656}}{\text{.71}}\,{\text{c}}{{\text{m}}^3}$
Now, we have to covert ${\text{c}}{{\text{m}}^{\text{3}}}$ to l. We know that, $1\,{\text{l}} = 1000\,{\text{c}}{{\text{m}}^{\text{3}}}$.
So, $656.71\,{\text{c}}{{\text{m}}^3} = \dfrac{{656.71}} {{1000}}\,\,{\text{l}} = 0.657\,\,{\text{l}}$

Hence, 0.657 l of propane is required to produce 3.9 g of ${\text{C}}{{\text{O}}_{\text{2}}}$.

Note: It is to be noted that Combustion reactions are very important kinds of chemical reactions. These reactions have an important impact on our daily life. When oxygen and fuel react, heat or fire and light are produced. Combustion of fossil fuels generates heat and this heat is used in operation of machines such as engines, boilers etc.