As you have 2 mol NaN3, you produce 3 mol N2 in the first reaction.

You have also produced 2 mol Na.

We know 10 Na produces 1 mol N2 in the secondary reaction.

However, as we only have 2 mol, we can only produce 2/10 mol N2.

Altogether we have: 3 + 0.2 = 3.2 mol N2

3.2 x Avagadro's constant (6.022 x 10^23) = 19.27 x 10^23 molecules

19.27 x 10^23 = 1.927 x 10^24 = 1.93 x 10^24 molecules

Hope that helps

You have also produced 2 mol Na.

We know 10 Na produces 1 mol N2 in the secondary reaction.

However, as we only have 2 mol, we can only produce 2/10 mol N2.

Altogether we have: 3 + 0.2 = 3.2 mol N2

3.2 x Avagadro's constant (6.022 x 10^23) = 19.27 x 10^23 molecules

19.27 x 10^23 = 1.927 x 10^24 = 1.93 x 10^24 molecules

Hope that helps

Why does the 3n2 react with the 10na and what do u do about the k in the second equation

Original post by Roxibox123

Why does the 3n2 react with the 10na and what do u do about the k in the second equation

No N2 reacts with Na at any point.

As we are focussed on the production of N2 only, we do not need to worry about K, we can just assume there is enough to react fully unless stated otherwise I believe.

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