Question #35c5e

1 Answer
Oct 30, 2016

Answer:

The volume of gas is 184 L.
#P_"N₂"# = 0.208 atm, and #P_"CO₂"# = 1.04 atm.

Explanation:

Your professor is an exceptional chemist if he can extract hydrogen ions by the electrolysis of water and ozonize #"CO"_2# in the Martian atmosphere to #"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

#"NO"_3^"-" + "CO" → "N"_2 + "CO"_2#

The half-reactions are:

#"2NO"_3^"-" + "12H"^"+" + "10e"^"-" → "N"_2 + "6H"_2"O"#
#5×["CO" + "H"_2"O" → "CO"_2 + "2H"^"+" + "2e"^"-"]#
#"2NO"_3^"-" + "5CO" + "2H"^"+" → "N"_2 + "5CO"_2 + "H"_2"O"#

We see that 2 mol of #"NO"_3^"-"# gives 6 mol of gas (1 mol of #"N"_2# and 5 mol of #"CO"_2#). The water will be a liquid under these conditions.

#"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"#

#PV = nRT#

#V = (nRT)/P#

#T = "17 °C" = "290.15 K"#
#"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_"N₂" = 1/6#, and #chi_"CO₂" = 5/6#.

#P_"N₂" = chi_"N₂"P_"tot" = 1/6 × "1.25 atm" = "0.208 atm"#

#P_"CO₂" = chi_"CO₂"P_"tot" = 5/6 × "1.25 atm" = "1.04 atm"#