The following equation represents the oxidation of ammonia,NH₃.
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
At the same temperature and pressure,what is the maximum volume of nitrogen monoxide that can be obtained from 7.55 × 10² L of ammonia and 7.55 × 10² L of oxygen?

Question

Quizplus · Accepted Answer

According to the balanced equation, 4 moles of NH3 will produce 4 moles of NO. In this case, we have 7.55 × 10^12 L of NH3 which is equivalent to 4 × 7.55 × 10^12 = 3.02 × 10^13 moles of NH3.

Similarly, we have 7.55 × 10^12 L of O2 which is equivalent to 5 × 7.55 × 10^12 = 3.78 × 10^13 moles of O2.

Now, we need to find out which reactant is the limiting one. To do this, we compare the mole ratio of NH3 and O2 in the equation, which is 4:5.

So, we can say that O2 is the limiting reactant because we have only 3.78 × 10^13 moles of O2 whereas we have 3.02 × 10^13 moles of NH3.

Now, we need to find out how many moles of NO will be produced from 3.78 × 10^13 moles of O2.

From the equation, we can see that 5 moles of O2 will produce 6 moles of H2O.

Therefore, 3.78×10^13 moles of O2 will produce (6/5) × 3.78×10^13 = 4.536×10^13 moles of H2O.

Next, we can see from the equation that 4 moles of NH3 will produce 4 moles of NO.

Therefore, 3.02 × 10^13 moles of NH3 will produce 3.02 × 10^13 moles of NO.

So, the amount of NO produced will be limited by the number of moles of O2 available, which is 4.536×10^13 moles of H2O.

Now, we need to find the volume of NO that corresponds to 4.536×10^13 moles of NO. To do this, we use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Assuming that the temperature and pressure are constant, we can write:

V = (nNO × RT) / P

where nNO is the number of moles of NO, R is the gas constant, and T is the temperature.

We can simplify this to:

V = (nNO / P) × (RT)

Plugging in the values, we get:

V = (4.536×10^13 / 1) × (0.0821 × 298) = 1.08 × 10^13 L

Therefore, the maximum volume of NO that can be obtained is 1.08 × 10^13 L, which is closest to option C, 6.04 × 10^12 L.

The Following Equation Represents the Oxidation of Ammonia,NH3

The Following Equation Represents the Oxidation of Ammonia,NH₃