497,205 views
45 votes
45 votes
N₂(g) + 3H₂(g) → 2NH3(g)

224 L of nitrogen reacts with
excess hydrogen at 2773 K and
95.0 atm. How many moles of
nitrogen react?

N₂(g) + 3H₂(g) → 2NH3(g) 224 L of nitrogen reacts with excess hydrogen at 2773 K and-example-1
User Midhun Pottammal
by
3.0k points

2 Answers

26 votes
26 votes

Final answer:

To find how many moles of nitrogen gas react, the ideal gas law was used. After rearranging the equation to solve for the number of moles and plugging in the given values of pressure, volume, and temperature, it was found that 9.30 moles of nitrogen gas react.

Step-by-step explanation:

The student asked how many moles of nitrogen react when 224 L of nitrogen gas is involved in the reaction with excess hydrogen to form ammonia, under the conditions of 2773 K and 95.0 atm. To calculate this, we can use the ideal gas law: PV = nRT, where P is the pressure (95.0 atm), V is the volume of the gas (224 L), n is the number of moles of gas, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature (2773 K).

First, rearrange the ideal gas law to solve for n: n = PV/(RT). Substituting the known values into this equation, we get n = (95.0 atm × 224 L) / (0.0821 L·atm/(mol·K) × 2773 K). Performing the calculation: n = 21120 atm·L / (227.173 mol·K) = 9.30 moles. Therefore, 9.30 moles of nitrogen gas react under the given conditions.

User Balinti
by
2.8k points
27 votes
27 votes

Answer:

93.5

Step-by-step explanation:

User Brian Beckett
by
2.9k points