# Calculate the mole of carbon dioxide when 11.2L of methane atSTP is ignited in excess of air?please explain?

Jun 4, 2018

n("CO"_2)=0.493 color(white)(l)"mol"

#### Explanation:

Steps:

1. Determine the number of moles of methane molecules in $11.2 \textcolor{w h i t e}{l} \text{L}$ of the gas at $\text{STP}$.
2. Calculate the number of moles of ${\text{CO}}_{2}$ produced with the stoichiometric relationship in the reaction between methane and excess oxygen.

The IUPAC defines $\text{STP}$ as a set of conditions under which

• Temperature $T = 273.15 \textcolor{w h i t e}{l} \text{K}$, and
• Pressure $P = {10}^{5} \textcolor{w h i t e}{l} \text{Pa}$. [1]

By the ideal gas law

$P \cdot V = n \cdot R \cdot T$

$n = \frac{P \cdot V}{R \cdot T}$

However, ideal gas constant $R$ is typically given in SI units as

$R = 8.314 \textcolor{w h i t e}{l} {\text{N" * "m" * "K}}^{- 1}$;

Some unit conversions would thus be necessary for the values given in this question.

$1 \textcolor{w h i t e}{l} {\text{L" = 1 color(white)(l) "dm}}^{3}$
$V = 11.2 \textcolor{w h i t e}{l} {\text{dm"^3=11.2*10^(-3) color(white)(l) "m"^(3)=1.12 xx10^(-2) color(white)(l) "m}}^{3}$

$11.2 \textcolor{w h i t e}{l} \text{L}$ of gas would, thus, correspond to

n=(10^5 color(white)(l) "N" * color(red)(cancel(color(black)("m"^(-2)))) * 1.12 xx 10^(-2) color(white)(l) color(red)(cancel(color(black)("m"^(3))))) / (8.314 color(white)(l) "N" * color(red)(cancel(color(black)("m"))) * "K"^(-1) * 273.15 color(white)(l) "K")=0.493 color(white)(l) "mol" of molecules.

Methane combusts in the air to produce carbon dioxide ${\text{CO}}_{2}$ and water $\text{H"_2"O}$ when oxygen is in excess. As seen in the balanced equation,

$\textcolor{\mathrm{da} r k g r e e n}{1} \textcolor{w h i t e}{l} \text{CH"_4 (g) + 2 color(white)(l) "O"_2 (g) to color(darkgreen)(1) color(white)(l) "CO"_2 (g) + 2 color(white)(l) "H"_2"O} \left(g\right)$,

(n("CO"_2))=(n("CH"_4))=0.493 color(white)(l) "mol".

Note, that some textbooks might assume the IUPAC definition of $\text{STP}$ before the year 1982, where

$P = 1 \textcolor{w h i t e}{l} \text{atm"= 1.01325 * 10^5 color(white)(l) "Pa}$

is deemed the pressure under standard conditions. See what value would you get for $n \left({\text{CO}}_{2}\right)$ given $P = 1.01325 \cdot {10}^{5} \textcolor{w h i t e}{l} \text{Pa}$.