Question

When 35.6L of ammonia gas and 40.5L of oxygen gas at burn, nitrogen monoxide gas and liquid water are produced. How many grams of nitrogen monoxide are present if the reaction takes place under standard pressure and temperature?

If we consider this reaction to take place in a cylinder with a moveable piston, then at the end of the reaction the piston has...

A. remained in the same position

B. Moved down

C. Moved up

D. Cannot be determined based on the available information\

Answer 1

At STP 1 mol of gas occupies 22.4 L.

Moles of ammonia = 35.6 /22.4 = 1.59 mol

Moles of oxygen = 40.5 /22.4 = 1.80 mol

2 NH₃ + 2 O₂ N₂O + 3 H₂O

From the balanced equation, it is clear that 2 moles of ammonia react with 2 moles of oxygen to give 1 mol of N₂O and 3 moles of water.

In this reaction, mole ratio of NH₃ : O₂ = 1.59 : 1.80

                                                             = 1.00 : 1.13

Required mole ratio of NH₃ : O₂ = 2:2

                                                   = 1:1

Hence, NH₃ is the limiting reagent in this reaction.

Now, 2 mol of NH₃ give 1 mol of N₂O

Therefore, 1.59 mol of NH₃ will give = 1.59/2 = 0.795 mol of N₂O

Molar mass of N₂O = 44 g/mol

Hence, the mass of the N₂O produced = 44 x 0.795 = 35.0 g

Thus, the grams of nitrogen monoxide are present if the reaction takes place under STP = 35.0 g

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Total moles of gaseous substance before the reaction = (1.59 + 1.80) = 3.39 mol

Total moles of gaseous substance after the reaction = unreacted mol of O₂ + mol of N₂O

                                                                                   = (1.80 - 1.59) + 0.795

                                                                                   = 0.21 + 0.795

                                                                                  = 1.01 mol

Thus, the moles of gasoeus substance is decreased after the reaction. Hence, the volume is also decreased. Therefore, the piston has moved down at the end of the reaction.