3O2(g)Â Â Â Â Â Â
+Â Â Â Â 4NH3(g)
------------------>Â Â 6H2O(g) + 2N2(g)
0.428Â Â Â Â Â
0.773Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â
0
0.428 - 3x        Â
0.773-4x                               Â
6x          Â
2x
If 0.094 moles of O2(g) react
3x = 0.094Â Â
x = 0.03133
moles of NH3 must react = 4x = 4 x 0.03133
                                     Â
= 0.125 moles
moles of NH3 must react = 0.125 moles
moles of H2O formed = 6x = 6 x 0.03133 = 0.188
moles
moles of N2 formed = 2x = 2 x 0.03133 = 0.0627
moles
moles of O2 remain = 0.428 - 3x = 0.334
moles of NH3 remain = 0.773-4x = 0.648