Question

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?

Answer 1

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.