How many molecules (not moles) of NH3 are produced from 8.97×10-4 g of H2

Question

How many molecules (not moles) of NH3 are produced from 8.97×10−4 g of H2

Answer 1

To answer this question you need first to write a balanced reaction of formation of NH₃ from H₂ and N₂:

                       

                                     N₂(g) + 3 H₂(g) → 2 NH₃(g)

 

Then calculate the moles of H₂ from the mass of H₂:

8.97 x 10^-4 g H₂ x (1 mol H₂/2.016 g H₂) = 4.45 x 10^-4 mol H₂

 

Based on the stoichimetry of the reaction:

 

 

4.45 x 10^-4 mol H₂ x (2 mol NH₃/3 mol H₂) = 2.97 x 10^-4 mol NH₃   

 

Then the molecules of NH3(g) can be calculated from Avogadro number:

 

2.97 x 10^-4 mol NH₃ x (6.022 x 10²³ molecules/ 1 mol NH₃) = 1.79 x 10²⁰ molecules NH₃ 

Answer 2

I'm presuming this is from the following balance reaction:

 

N₂ + 3H₂ → 2NH₃

 

As you can see from the reaction, 3 mol H₂ are used to produce 2 mol NH₃.

We will also need the molar mass for both H₂ and NH₃.

 

MW (H₂) = 2.02 g/mol

MW (NH₃) = 17.03 g/mol

 

We will also need to know Avogadro's number:

N_A = 6.022 x 1023 molecules or atoms

 

We will use dimensional analysis to arrive at the final answer.

 

8.97 x 10^-4 g H₂ x (1 mol H₂ / 2.02 g H₂) x (2 mol NH₃ / 3 mol H₂) x (6.022 x 10²³ molecules NH₃ / 1 mol NH₃)

 

Answer: ??? 

(Calculate)