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

queen_honey_blossom answered 1 year ago

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