nN2:nH2=2:3=> N2 dư, H% tính theo H2
dX/H2=\(\dfrac{28.2+3.2}{\left(2+3\right).2}\)= 6,2
Có \(\dfrac{d_{\dfrac{X}{H2}}}{d_{\dfrac{Y}{H2}}}=\dfrac{n_Y}{n_X}\) => \(\dfrac{6,2}{7,75}=\dfrac{n_Y}{2+3}\) => nY= 4
N2 + 3H2 ⇌ 2NH3
2 3
x 3x 2x
2-x 3-3x 2x
ngiảm=nNH3=nX-nY= 5-4=1=>x=0,5
=> nH2 phản ứng= 3.0,5=1,5
H%=\(\dfrac{1,5.100}{3}\)=50%