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mNO=30u (m2)
4NH3 + 3O2 ---> 2N2 + 6H2O
4NH3 + 3O2 ---> 4NO + 6H2O
m1=28
m2=30
X1=x%
X2=(100-x)%
Korzystam ze wzoru na średnią masę atomową:
m at. = m1*X1 + m2*X2 / 100%
28,7 = 28x + 30(100-x) / 100%
2870 = 28x + 3000 - 30x
2x = 130
x = 65
X1=65%
X2=35%
4mole NH3 ------ 2mole N2
xmoli NH3 ------- 0,65mola N2
x = 1,3mola NH3
4mole NH3 ------ 4mole NO
xmoli NH3 ------- 0,35mola NO
x = 0,35mola NH3
nNH3 = 0,35 /(1,3+0,35)
nNH3 = 0,212mola