4NH3 + 5O2 = 4NO + 6H2O How many moles of NO are produced if 6.96 grams of NH3 react with an excess amount of oxygen?

Question

Grateful for any help! If you could explain the work that'd be helpful as well!

Answer 1

Start with a balanced equation which you already have.

4NH3 + 5O2 = 4NO + 6H2O

Now ask yourself, "what do I want to find". You want to know how many moles of NO are produced.

What are you given? You are given that you start with 6.96 g NH3 and excess O2.

Now you need the correct conversion factor. The conversion factor is a fraction with the numerator containing the quantity you are trying to find, and the denominator containing what you are given.

4 moles NO / 4 x mol mass of NH3 Note that the numerator is in moles of NO which is the coefficient of NO in the balanced equation. The denominator includes the coefficient for NH3 x mol. mass of NH3. It should be this way whenever a substance is given in grams. Putting numbers in the factor gives you

4 moles NO / 4 x 17.031 g /mole = 0.0587 moles NO per gram of NH3. Now multiply the factor result by the number of grams NH3 you were given.

0.0587 moles NO / g NH3 x 6.96 g NH3 = 0.409 g NO produced.

This is probably not how it is taught, but I find it more straightforward.

Answer 2

Good afternoon. This problem can be solved in two steps:

Step 1: convert grams of NH₃ to moles of NH₃

Step 2: convert moles of NH₃ to moles of NO using the coefficients in the balanced equation

(6.96g NH₃) X (1 mol NH₃ / 17.031g) = 0.409 mol of NH₃

(0.409 mol NH₃) X (4 mol NO / 4 mol NH₃) = 0.409 mol NO

Because NH₃ and NO both have a coefficient of 4 in the balanced equation, the number of moles are the same for those particular compounds.