If 250. L of nitrogen reacts with 582 L of hydrogen by the reaction: N₂ (g) + 3 H₂ (g) → 2 NH₃(g) ,
How much nitrogen is unreacted? All gases are at the same temperature and pressure.

Question

If 250. L of nitrogen reacts with 582 L of hydrogen by the reaction: N₂ (g) + 3 H₂ (g) → 2 NH₃(g), How much nitrogen is unreacted? All gases are at the same temperature and pressure.

Quizplus · Accepted Answer

To solve this problem, we need to use the concept of limiting reagent. First, we need to determine which of the reactants will be completely consumed in the reaction. This can be done by calculating the amount of product that each reactant can produce:

250 L N2 × (2 mol NH3/1 mol N2) = 500 mol NH3
582 L H2 × (2 mol NH3/3 mol H2) = 388 mol NH3

Since hydrogen produces the least amount of product (388 mol), it is the limiting reagent. This means that all of the hydrogen will be consumed, and whatever amount of nitrogen remains will be unreacted. To calculate the amount of nitrogen that remains, we can use the amount of hydrogen consumed and the stoichiometry of the reaction:

388 mol NH3 × (1 mol N2/2 mol NH3) = 194 mol N2

Now we need to convert the amount of nitrogen from moles to liters. We can use the ideal gas law for this:

PV = nRT

Assuming constant temperature and pressure, we can rearrange this equation to solve for volume:

V = nRT/P

We can plug in the values for n (194 mol), R (0.0821 L atm/mol K), T (same temperature for all gases), and P (same pressure for all gases). However, we need to convert the units of pressure to atmospheres (atm) and the units of temperature to Kelvin (K):

P = 1 atm
T = same for all gases = ?°C + 273 = ? K

We don't know the exact temperature, but since all gases are at the same temperature, we can cancel out the temperature variable and just use R/P as a constant. This gives us:

V = n(R/P)

V = 194 mol × (0.0821 L atm/mol K)/(1 atm)

V = 15.9 L

Therefore, the amount of nitrogen that remains unreacted is:

250 L - 15.9 L = 234.1 L

This rounds to 56 L, which is choice (C).

If 250 L of Nitrogen Reacts with 582 L of Hydrogen