Ammonia is often formed by reacting nitrogen and hydrogen gases. How many liters of ammonia gas can be formed from 23.7 L of hydrogen gas at 93.0°C and a pressure of 38.9 kPa?
N2 + 3H2 ==> 2NH3
If you want the volume of NH3 at the same conditions, it is just 23.7 L x (2 mols NH3/3 mols H2) = ?
If you want it at STP, use PV = nRT and convert.
To solve this problem, we can use the ideal gas law, which states that 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 in Kelvin.
First, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 93.0 + 273.15
T(K) = 366.15 K
Next, we need to calculate the number of moles of hydrogen gas (H2) using the ideal gas law:
PV = nRT
P = 38.9 kPa
V = 23.7 L
R = 8.314 J/(mol·K) [gas constant]
Now, let's solve for n:
n = PV / RT
n = (38.9 kPa * 23.7 L) / (8.314 J/(mol·K) * 366.15 K)
n ≈ 2.547 moles
Since the balanced equation for the reaction is:
N2 + 3H2 → 2NH3
We can see that the ratio of moles of H2 to moles of NH3 is 3:2. Therefore, for every 3 moles of H2, we get 2 moles of NH3.
So, the number of moles of NH3 obtained will be:
(2/3) * 2.547 moles ≈ 1.698 moles
Finally, we can calculate the volume of ammonia gas (NH3) using the ideal gas law:
PV = nRT
P = 38.9 kPa [pressure of ammonia]
V = ? [volume of ammonia]
n = 1.698 moles [number of moles of ammonia]
R = 8.314 J/(mol·K) [gas constant]
T = 366.15 K [temperature]
Now, let's solve for V:
V = nRT / P
V = (1.698 moles * 8.314 J/(mol·K) * 366.15 K) / (38.9 kPa)
V ≈ 4.86 L
Therefore, approximately 4.86 L of ammonia gas can be formed from 23.7 L of hydrogen gas at 93.0°C and a pressure of 38.9 kPa.
To find out how many liters of ammonia gas can be formed, we need to use the ideal gas law equation. The ideal gas law equation is given by: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, we need to convert the given temperature from Celsius to Kelvin by adding 273.15. So, 93.0°C + 273.15 = 366.15 K.
Next, we need to find the number of moles of hydrogen gas. We can use the ideal gas law equation to find the number of moles. Rearranging the equation, we have n = PV / RT. Let's calculate it.
Given:
- Hydrogen gas volume (V) = 23.7 L
- Hydrogen gas pressure (P) = 38.9 kPa
- Hydrogen gas temperature (T) = 366.15 K
- R is the ideal gas constant = 8.314 L·kPa·K⁻¹·mol⁻¹
n = (38.9 kPa * 23.7 L) / (8.314 L·kPa·K⁻¹·mol⁻¹ * 366.15 K)
n = 218.12 mol
According to the balanced chemical equation, the reaction between nitrogen and hydrogen gases produces 2 moles of ammonia gas for every 3 moles of hydrogen gas.
Therefore, if we have 218.12 moles of hydrogen gas, we will produce (2/3) * 218.12 moles of ammonia gas.
Let's calculate the number of moles of ammonia gas:
moles of ammonia gas = (2/3) * 218.12
moles of ammonia gas = 145.41 mol
Finally, to find the volume of ammonia gas in liters, we use the ideal gas law equation again, rearranging it for volume:
V = nRT / P
Given:
- Ammonia gas moles (n) = 145.41 mol
- Ammonia gas temperature (T) = 366.15 K
- Ammonia gas pressure (P) = ? (not provided)
Since the pressure of ammonia gas is not provided, it is not possible to calculate the volume without that information.