Question

Nitrogen forms 3 oxides, NO, NO2 and N20. Calculate the percentage of nitrogen in each oxide

Could someone please work this out for me and possible explain.

Answer 1

Explanation:

There are many chemical reactions that can be used to produce nitrogen monoxide (NO), also called nitric oxide. Nitric oxide - Wikipedia

The simplest reaction occurs between N2 and O2 in the atmosphere, when the temperature reaches >2000°C due to lightening.

This reaction is:

N2(g) + O2(g) → 2NO

To calculate the mass of N2 required to react with 32 g of O2, the process will be:

mass O2 → mole O2 → mol N2 → mass N2

Convert 32 g O2 to moles O2.

n = m/M

n = mole = ?

m = mass(g)

M = molar mass (g/mol) = (2 × 16 g/mol O) = 32 g/mol O2

n O2 = 32 g O2/32 g O2/1 mol O2 = 1 mol O2

Convert mol O2 to mol N2.

Multiply mol O2 by the mole ratio between O2 and N2 in the balanced equation, so that mole O2 cancels, leaving mole N2.

1 mol O2 × 1 mol N2/1 mol O2 = 1 mol N2

Calculate mass N2.

Rearrange n = m/M to isolate m.

m = n × M

M N2 = (2 × 14 g/mol N) = 28 g/mol N2

m N2 = 1 mol N2 × 28 g N2/1 mol N2 = 28 g N2

28 g of N2 are needed to react with 32 g of O2 to form NO.

Note: Since the mole ratio between N2 and O2 is 1:1, once we determined that we had 1 mol O2, we could have deduced that there was also 1 mol N2, which we would then multiply by its molar mass to get the mass of N2.

If you already knew the molar mass of O2 and N2, you could have solved this question in your head very quickly if you did not need to show the math.

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