Maximum theoretical yield?

Question

What is the maximum theoretical yield of ammonia when 280g of nitrogen is reacted with 60g of hydrogen?
H=1
N=7
Please help! I am really stuck!

Thanks!

Answer 1

Your atomic weight for nitrogen is wrong. You are using the atomic number (7) and not the weight (14).
For 1 molecule of ammonia (NH3) you need 1 mole of nitrogen and 3 moles of hydrogen.
280 g of nitrogen are 20 moles, 60 g of hydrogen are 60 moles. With the nitrogen you could form 20 moles of ammonia, and with the hydrogen 60/3 = 20 moles too.
The theoretical amount is 280 + 60 = 340 g of ammonia (or 20 moles).

Answer 2

N + 3H -> NH3 is a quick lazy formula for what you want. Use N=14, though.

1g of H + 42 g of N -> 43 g of NH3

Notice that 60 g hydrogen needs 60 X 42 g of nitrogen i.e. more than 2,400 g.

Nitrogen is limiting in the problem given.

I have no calculator to hand so:

210 g nitrogen gives 43X5 = 215 g ammonia
42 g give 43 g ammonia
21 g nitrogen give 21.5 g ammonia
7 g nitrogen give 7.17 g ammonia
so
280 g nitrogen give 215 + 43 + 21.5 + 7.17 g ammonia.

Answer 3

N2 + 3H2 = 2NH3
Note the molar ratios are 1:3::2

Mr(N2) = 28
Mr(H2) = 2
Mr(NH3) = 17

You have quoted the atomic numbers, it is the atomic masses that are needed.
Next calculate the moles of reactants using the equation:-
moles(n) = mass/g / Mr
moles(N2) = 280g/28 = 10 moles(N2)
moles(H2) = 60g/2 = 30 moles(H2)
N2 : H2 :: 10 : 30 :: 1 : 3

So the masses of gases react completely with one another without any gas being in excess.
NH3 : 2 : 20
moles(NH3) = mass(NH3) / Mr(NH3)
mass(NH3) = moles(NH3) x Mr(NH3)
mass(NH3) = 20 x 17 = 340 g (100% yield as no gas is in excess).