Ammonia and oxygen produce nitrogen dioxide and water. What volume of nitrogen dioxide gas will be produced if 30.5 grams of ammonia is reacted with excess oxygen?

1 Answer
Apr 1, 2016

Answer:

40.7 L of #"NO"_2# will be produced at STP.

Explanation:

There are four steps involved in this stoichiometry problem:

Step 1. Write the balanced chemical equation.

#"4NH"_3 + "7O"_2 → "4NO"_2 + "6H"_2"O"#

2. Convert grams of #"NH"_3# to moles of #"NH"_3#

The molar mass of #"NH"_3# is 17.03 g/mol.

#"moles of NH"_3 = 30.5 color(red)(cancel(color(black)("g NH"_3))) × ("1 mol NH"_3)/(17.03 color(red)(cancel(color(black)("g NH_3"_3)))) = "1.791 mol NH"_3#

Step 3. Convert moles of #"NH"_3# to moles of #"NO"_2#**

The molar ratio of #"NO"_2# to #"NH"_3# is #"4 mol NO"_2:"4 mol NH"_3#.

#1.791 color(red)(cancel(color(black)("mol NH"_3))) × ("4 mol NO"_2)/(4 color(red)(cancel(color(black)("mol NH"_3)))) = "1.791 mol NO"_2#

4. Use the Ideal Gas Law to calculate the volume of #"NO"_2#.

The Ideal Gas Law is

#color(blue)(|bar(ul(color(white)(a/a) PV = nRTcolor(white)(a/a)|)))#, where

  • #P# = the pressure of the gas,
  • #V# = the volume of the gas,
  • #n# = the number of moles of the gas,
  • #R# = the universal gas constant
  • #T# = the temperature of the gas

You don't give the pressure or temperature of the gas, so I will assume STP (100 kPa and 0 °C).

We can rearrange the Ideal Gas Law to get

#color(blue)(|bar(ul(color(white)(a/a)V= (nRT)/Pcolor(white)(a/a)|)))" "#

#V = (1.791 color(red)(cancel(color(black)("mol"))) ×8.314 color(red)(cancel(color(black)("kPa")))"·L·"color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(100 color(red)(cancel(color(black)("kPa")))) = "40.7 L"#

The volume of #"NO"_2# is 40.7 L.

Here's a useful video on mass-volume calculations.