If 6.52l of h2 react with excess nitrogen what volume of nh3 is produced?

The volume of NH₃ produced is 3.6 L

Explanation:

Let's make the reaction:

N₂ (g)  +  3H₂ (g)  ⇄  2NH₃ (g)

We have the volume of H₂ which has reacted, but we don't have the mass (neither the moles). We must apply density of H₂ to know this data:

δH₂ (g) =  0.071 g/l

density = mass / volume

0.071 g/L = mass / 6.52L

0.071 g/L . 6.52L = mass → 0.463 g

Molar mass of H₂ = 2 g/m

Mass / Molar mass = moles

0.463 g /2g/m = 0.231 moles

In the reaction ratio between hydrogen and ammonia is 3:2

3 moles of H₂ 2 moles of ammonia

0.231 moles H₂   (0.231 .2)/3 = 0.154 moles

These are the moles of ammonia which are produced, so we have to convert this to mass, to apply NH₃ 's density.

moles . molar mass = mass

Molar mass NH₃ = 17 g/m

0.154 m . 17g/m = 2.62 g

NH₃ density = 0,00073 g/ml

NH₃ density = NH₃ mass / NH₃ volume

NH₃ volume =  NH₃ mass / NH₃ density

NH₃ volume = 2.62 g /0,00073 g/ml

NH₃ volume = 3593.4 mL

3593.4 mL /1000 ≅ 3.6L

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