Asked by wallace
In a reaction of ammonia production (NH3), 360g of nitrogen gas (N2) was reacted, generating a yield of 20%. What was the mass obtained from ammonia?
Answers
Answered by
Anonymous
I guess this is the Haber process:
N2 + 3H2 ---> 2NH3
360g*(mole/28.0134g)= moles of N2
moles of N2*(2 moles of NH3/1 mole of N2)= moles of NH3
Moles of NH3*(17.03g/mole)= 100% yield
0.20*(100% yield)=NH3 produced
N2 + 3H2 ---> 2NH3
360g*(mole/28.0134g)= moles of N2
moles of N2*(2 moles of NH3/1 mole of N2)= moles of NH3
Moles of NH3*(17.03g/mole)= 100% yield
0.20*(100% yield)=NH3 produced
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