how many grams of ammonia formed during the reaction with 100g of hydrogen and 100g of nitrogen?

Question

n2+3h2 -> 2nh3 .. so confused? 

Answer 1

First, you really should use the correct way of writing chemical formulae.  That is...

N₂ + 3H₂ ==> 2NH₃ 

Having said that, now on to how to solve this problem...

Note the stoichiometry which tells you that it takes 1 mol of N₂ to react with 3 moles of H₂ to prepare 2 moles NH₃.

If we now find how many moles of N2 and H2 we have, we can then find the LIMITING REACTANT.

Moles N₂ = 100 g N₂ x 1 mole N₂/28 g = 3.57 moles N₂

Moles H₂ = 100 g H₂ x 1 mole H₂/2 g = 50 moles H₂

Do we have 3x as much H₂ as N₂ as indicated by the stoichiometry?  Yes, we have much more than 3x as much H₂.  This means that N₂ must be limiting, as it will run out way before the H₂ is all used up.

Now that we know N₂ is limiting, we go back and look at the stoichiometric relationship between N₂ and NH₃ and we see that 1 mole of N₂ produces 2 moles of NH₃.  We can now calculate moles of NH₃ that will appear.

Moles of NH₃ formed = 3.57 moles N₂ x 2 moles NH₃/1 mole N₂ = 7.14 moles NH₃

Converting moles to grams we have 7.14 moles NH₃ x 17 g NH₃/mole = 121 grams NH₃ produced (to 3 significant figures).  NOTE: the 100 g as written in the question has only 1 significant figure and thus the answer should more correctly by 100 grams NH₃, but I assumed that maybe it was 100. grams of each gas.  Maybe not such a good assumption on my part).