In one production cycle, 28 x 10^3 kg of N2(g) reacts with 6 x 10^3 kg of H2(g) .
8.5 x 10^3 kg of NH3 (g) are produced. What is the yield?

I have no idea how to calculate the yield. I feel like it should be something really simple though...?

Question

In one production cycle, 28 x 10^3 kg of N2(g) reacts with 6 x 10^3 kg of H2(g) .
8.5 x 10^3 kg of NH3 (g) are produced. What is the yield?

I have no idea how to calculate the yield. I feel like it should be something really simple though...?

butterflydreamer · Answer

do i just add the mass of N2 and the mass of H2 together..
Then divide the mass of NH3 produced by the mass of (N2 + H2) ? LOL just a thought

butterflydreamer · Answer

If so, it would be 25% yield right?

syedmohammed98 · Answer

ahhh spent like 20 minutes oh this question ...m confused too.... um u may be correct ...but m not sure...i mean all i was doing is finding their number of moles and stuff... btw the formula for finding a yield of a reaction is   (actual yield / calculate yield) * 100

syedmohammed98 · Answer

calculated yield*

butterflydreamer · Answer

hmm interesting.
Tbh.. i think it's just 25% xD 
And thaaank you for trying and telling me the formula :)

syedmohammed98 · Answer

You're Welcome
actually i was thinking that 25% is too low.. i mean the scientist was not good at performing experiment might be using cellphone during the experiment .so didn't controlled the other factors .lol...

butterflydreamer · Answer

lmaooo true... but i guess since they haven't mentioned the conditions like temp, pressure etc.. i guess we have to assume that the scientist is performing it in the optimal conditions?

syedmohammed98 · Answer

yup.....damaged apparatus can also be a reason for less yield

aaronq · Answer

You're supposed to find the theoretical yield - using the limiting reactant - then use:

$\sf \% yield=\dfrac{|theoretical-experimental|}{experimental}*100\%$