3O2(g) + 4NH3(g) 6H2O(g) + 2N2(g) Initially (before any reaction occurs) a 1.00 liter reaction vessel...

3.9 Ratings (705 Votes)

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 - 3xÂ Â Â Â Â Â Â Â Â 0.773-4xÂ 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 6xÂ Â Â Â Â Â Â Â Â Â 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 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 Â 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

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