Question #ed921

1 Answer
Jun 23, 2017

Answer:

The reaction produces 2.84 L of #"CO"_2#.

Explanation:

Step 1. Start with the balanced equation

#"2C"_2"H"_2 + "5O"_2 → "4CO"_2 + "2H"_2"O"#

Step 2. Calculate the moles of #"O"_2#

#"Moles of O"_2 = 5.00 color(red)(cancel(color(black)("g O"_2))) × "1 mol O"_2/(32.00 color(red)(cancel(color(black)("g O" _2)))) = "0.1562 mol O"_2#

Step 3. Calculate the moles of #"CO"_2#

#"Moles of CO"_2 = 0.1562 color(red)(cancel(color(black)("mol O"_2))) × "4 mol CO"_2/(5 color(red)(cancel(color(black)("mol O"_2)))) = "0.1250 mol O"_2#

Step 4. Use the Ideal Gas Law to calculate the moles of #"CO"_2#

The Ideal Gas Law is

#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#

where

  • #p# is the pressure
  • #V# is the volume
  • #n# is the number of moles
  • #R# is the gas constant
  • #T# is the temperature

We can rearrange the Ideal Gas Law to get

#V = (nRT)/p#

STP is 1 bar and 0 °C.

#V = (0.1250 color(red)(cancel(color(black)("mol"))) × "0.083 14" color(red)(cancel(color(black)("bar")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "2.84 L"#