Question

LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane $\left( {60\% } \right)$ and butane $\left( {40\% } \right)$. If $10{\text{ litre}}$ of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reactions can be represented as:
${{\text{C}}_{\text{3}}}{{\text{H}}_{{\text{8}}\left( {\text{g}} \right)}} + {\text{5}}{{\text{O}}_{{\text{2}}\left( {\text{g}} \right)}} \to {\text{3C}}{{\text{O}}_{{\text{2}}\left( {\text{g}} \right)}} + {\text{4}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( {\text{g}} \right)}}$
${\text{2}}{{\text{C}}_{\text{4}}}{{\text{H}}_{10\left( {\text{g}} \right)}} + {\text{13}}{{\text{O}}_{{\text{2}}\left( {\text{g}} \right)}} \to {\text{8C}}{{\text{O}}_{{\text{2}}\left( {\text{g}} \right)}} + {\text{10}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( {\text{g}} \right)}}$
Calculate the percentage of nitrogen and oxygen in ammonium nitrate. [Relative molecular mass of ammonium nitrate is 80, ${\text{H}} = 1$, ${\text{N}} = 14$, ${\text{O}} = 16$]

Answer 1

To solve the first part first calculate the volumes of propane and butane. Then using the given combustion reactions, calculate the volume of carbon dioxide released by combustion of each propane and butane. In the second part, we must know the formula for ammonium nitrate.

Complete solution:
We are given that LPG are marketed including a mixture of propane $\left( {60\% } \right)$ and butane $\left( {40\% } \right)$. The total volume of the mixture is $10{\text{ litre}}$. Thus,
Volume of propane $= \dfrac{{60}}{{100}} \times 10 = 6{\text{ litre}}$
Volume of butane $= \dfrac{{40}}{{100}} \times 10 = 4{\text{ litre}}$
Consider the combustion reaction,
${{\text{C}}_{\text{3}}}{{\text{H}}_{{\text{8}}\left( {\text{g}} \right)}} + {\text{5}}{{\text{O}}_{{\text{2}}\left( {\text{g}} \right)}} \to {\text{3C}}{{\text{O}}_{{\text{2}}\left( {\text{g}} \right)}} + {\text{4}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( {\text{g}} \right)}}$
From the reaction, we can see that one volume of propane gives three volumes of carbon dioxide. Thus, the volume of carbon dioxide given by six volumes of propane is,
Volume of carbon dioxide $= 6{\text{ litre }}{{\text{C}}_3}{{\text{H}}_8} \times \dfrac{{3{\text{ litre C}}{{\text{O}}_2}}}{{1{\text{ litre }}{{\text{C}}_3}{{\text{H}}_8}}} = 18{\text{ litre C}}{{\text{O}}_2}$
Consider the combustion reaction,
${\text{2}}{{\text{C}}_{\text{4}}}{{\text{H}}_{10\left( {\text{g}} \right)}} + {\text{13}}{{\text{O}}_{{\text{2}}\left( {\text{g}} \right)}} \to {\text{8C}}{{\text{O}}_{{\text{2}}\left( {\text{g}} \right)}} + {\text{10}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( {\text{g}} \right)}}$
From the reaction, we can see that two volumes of butane give eight volumes of carbon dioxide. Thus, the volume of carbon dioxide given by four volumes of butane is,
Volume of carbon dioxide $= 4{\text{ litre }}{{\text{C}}_4}{{\text{H}}_{10}} \times \dfrac{{8{\text{ litre C}}{{\text{O}}_2}}}{{2{\text{ litre }}{{\text{C}}_4}{{\text{H}}_{10}}}} = 16{\text{ litre C}}{{\text{O}}_2}$
Thus,
Total volume of carbon dioxide $= \left( {18 + 16} \right){\text{ litre}}$
Total volume of carbon dioxide $= 34{\text{ litre}}$
Thus, the total volume of carbon dioxide gas added to the atmosphere is $34{\text{ litre}}$.
We have to calculate the relative percentage of nitrogen and oxygen in ammonium nitrate. The molecular formula for ammonium nitrate is ${\text{N}}{{\text{H}}_{\text{4}}}{\text{N}}{{\text{O}}_{\text{3}}}$.
We know that the relative percentage of each element is the mass of a single element divided by mass of the entire compound. The value is represented as a percentage.
The molecular mass of ammonium nitrate is 80. There are two nitrogen atoms in ammonium nitrate. The relative molecular mass of nitrogen is 14. Thus,
Relative percentage of nitrogen $= \dfrac{{2 \times 14}}{{80}} \times 100 = 35\%$
There are three oxygen atoms in ammonium nitrate. The relative molecular mass of oxygen is 16. Thus,
Relative percentage of oxygen $= \dfrac{{3 \times 16}}{{80}} \times 100 = 60\%$

Thus, the relative percentage of nitrogen and oxygen in ammonium nitrate is $35\%$ and $60\%$ respectively.

Note: Every chemical compound is a combination of two or more elements. Percent composition of the chemical compound tells us the percent by mass of each element present in the chemical compound. The relative percentage of each element is the mass of a single element divided by the mass of the entire compound.