nN2 = 4.48/22.4=0.2 mol
nNH3 = 1.7/17 = 0.1 mol
N2 + 4H2 -to-> 2NH3
0.05___________0.1
%H = 0.05/0.2*100% = 25%
https://i.imgur.com/onoMP2A.jpg
N2+3H2--->2NH3
nN2= 0,2(mol)
n\(_{NH3}=\frac{1,7}{17}=0,1\left(mol\right)\)
Theo pthh
n\(_{NH3}=n_{N2}=0,2\left(mol\right)\)
Suy ra H%=100%
Ta cs :
\(n_{N2}=0,2\left(mol\right)\)
\(PTHH:N2+3H2\rightarrow2HN3\)
\(\Rightarrow nHN3=nH2=0,2\left(mol\right)\)
\(\Rightarrow H\%=100\%\)
