Question
In: Chemistry
3O2(g) + 4NH3(g) 6H2O(g) + 2N2(g) Initially (before any reaction occurs) a 1.00 liter reaction vessel...
3O2(g) + 4NH3(g) 6H2O(g) + 2N2(g) Initially (before any reaction occurs) a 1.00 liter reaction vessel at 400 °C contains 0.428 moles of O2(g) and 0.773 moles of NH3(g) and no water or nitrogen. Consider the following: If 0.094 moles of O2(g) react, how many moles of NH3(g) must react and how many moles of H2O(g) and N2(g) are formed? How many moles of O2(g), NH3(g), H2O(g) and N2(g) remain after completion of the reaction?
Solutions
Expert Solution
3O2(g) + 4NH3(g) ------------------> 6H2O(g) + 2N2(g)
0.428 0.773 0 0
0.428 - 3x 0.773-4x 6x 2x
If 0.094 moles of O2(g) react
3x = 0.094
x = 0.03133
moles of NH3 must react = 4x = 4 x 0.03133
= 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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