# Question 35c5e

Oct 30, 2016

The volume of gas is 184 L.
${P}_{\text{N₂}}$ = 0.208 atm, and ${P}_{\text{CO₂}}$ = 1.04 atm.

#### Explanation:

Your professor is an exceptional chemist if he can extract hydrogen ions by the electrolysis of water and ozonize ${\text{CO}}_{2}$ in the Martian atmosphere to $\text{CO}$ and create an explosion at 1.25 atm pressure.

We should send him on the next rocket to Mars.

I presume that you are expected to use the reaction

${\text{NO"_3^"-" + "CO" → "N"_2 + "CO}}_{2}$

The half-reactions are:

$\text{2NO"_3^"-" + "12H"^"+" + "10e"^"-" → "N"_2 + "6H"_2"O}$
5×["CO" + "H"_2"O" → "CO"_2 + "2H"^"+" + "2e"^"-"]
$\text{2NO"_3^"-" + "5CO" + "2H"^"+" → "N"_2 + "5CO"_2 + "H"_2"O}$

We see that 2 mol of $\text{NO"_3^"-}$ gives 6 mol of gas (1 mol of ${\text{N}}_{2}$ and 5 mol of ${\text{CO}}_{2}$). The water will be a liquid under these conditions.

$\text{Moles of gas" = 200.0 color(red)(cancel(color(black)("g NO"_3^"-"))) × (1 color(red)(cancel(color(black)("mol NO"_3^"-"))))/(62.00 color(red)(cancel(color(black)("g NO"_3^"-")))) × "6 mol gas"/(2 color(red)(cancel(color(black)("mol NO"_3^"-")))) = "9.677 mol gas}$

$P V = n R T$

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

$T = \text{17 °C" = "290.15 K}$
$\text{P = 1.25 atm}$

V = (nRT)/P = (9.677 color(red)(cancel(color(black)("mol"))) × "0.082 06" "L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) ×290.15 color(red)(cancel(color(black)("K"))))/(1.25 color(red)(cancel(color(black)("atm")))) = "184 L"#

${\chi}_{\text{N₂}} = \frac{1}{6}$, and ${\chi}_{\text{CO₂}} = \frac{5}{6}$.

${P}_{\text{N₂" = chi_"N₂"P_"tot" = 1/6 × "1.25 atm" = "0.208 atm}}$

${P}_{\text{CO₂" = chi_"CO₂"P_"tot" = 5/6 × "1.25 atm" = "1.04 atm}}$