Question a9f52

Dec 29, 2017

$\text{10.7 L}$

Explanation:

For starters, you know by looking at the balanced chemical equation

${\text{CH"_ (4(g)) + 2"O"_ (2(g)) -> "CO"_ (2(g)) + 2"H"_ 2"O}}_{\left(l\right)}$

that in order for the reaction to produce $1$ mole of carbon dioxide, it must consume $1$ mole of methane. This basically means that the number of moles of methane that take part in the reaction will be equal to the number of moles of carbon dioxide produced by the reaction.

As you know, the number of moles of methane can be calculated by using the molar mass of the compound, ${\text{16.04 g mol}}^{- 1}$.

n_ ("CH"_ 4) = (6.56 color(red)(cancel(color(black)("g"))))/(16.04 color(red)(cancel(color(black)("g")))"mol"^(-1)) = 4.65/16.04color(white)(.)"moles"

This means that the reaction will produce

n_( "CO"_ 2) = 6.56/16.04color(white)(.)"moles CO"_2#

At this point, all you have to do to figure out the volume occupied by the gas is to use the ideal gas law equation.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the initial pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the mixture
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Now, you should know that standard pressure is currently defined as a pressure of

$\text{100 kPa" = 100/101.325color(white)(.)"atm}$

and that the temperature of the gas must be converted to Kelvin.

$T \left[\text{K"] = t[""^@"C}\right] + 273.15$

Keep in mind that when you're converting from degrees Celsius to Kelvin, the temperature in Kelvin must be rounded to the same number of decimal places as the temperature in degrees Celsius.

In this case, you have

${40}^{\circ} \text{C" + 273.15 = "313 K}$

The result has three significant figures and nod decimal places because ${40}^{\circ} \text{C}$ has no decimal places.

Rearrange the ideal gas law equation to solve for the volume of the gas

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

and plug in your values to find

$V = \left(\frac{6.56}{16.04} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (40 + 273.15)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm}}}}\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{V = \text{10.7 L}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the mass of methane and for the absolute temperature of the gas.