What volume of nitrogen gas can be produced from he decomposition of 37.6 L of ammonia with both gases measured at 725 degree celsius and 5.05 pressure?
In the reaction in 71 what total volume of gas at 667 celsius and 11.3 atm can be produced by the decomposition at 1.19 X 10 4L of ammonia at the same temperature and pressure?
The equation for the first one is
N2 + 3H2 => 2NH3
Refer to the CO problem below. This is not a LR problem but it's worked the same way.
For #2, what rhe reaction 71. I don't have a copy of yur text or notes but this is a decomposition of NH3 so the ci problem will work this one too.
Ok thanks.
What is the answer to the second one using CI
L N2 = 1.19E4L NH3 x (1/2) = ?
L H2 = 1.19E4L NH3 x (3/2) = ?
Total V at same T and P is the sum L N2 + L H2.
To answer these questions, we need to use the ideal gas law equation, which is:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature of the gas in Kelvin
We can rearrange the ideal gas law equation to solve for the volume of gas (V):
V = (nRT) / P
To get the number of moles (n) of a gas, we can use the equation:
n = PV / RT
Now, let's solve the first question:
1. Convert the given temperature from Celsius to Kelvin:
T = 725 + 273 = 998 K
2. Convert the given pressure to atm:
P = 5.05 atm
3. Use the given volume of ammonia to find the number of moles of ammonia:
n( NH3 ) = PV / RT
n( NH3 ) = (5.05 atm) * (37.6 L) / (0.0821 L·atm/(mol·K) * 998 K)
4. Since the stoichiometry of the reaction is 1 mole of NH3 yields 1 mole of nitrogen gas (N2), there will be the same number of moles of nitrogen gas as ammonia. Therefore, the number of moles of nitrogen gas is also:
n( N2 ) = n( NH3 )
5. Calculate the volume of nitrogen gas produced using the number of moles of nitrogen gas and the given temperature and pressure:
V( N2 ) = n( N2 ) * (RT / P)
V( N2 ) = n( NH3 ) * (RT / P)
V( N2 ) = (n( NH3 ) * (0.0821 L·atm/(mol·K) * 998 K)) / 5.05 atm
Now, let's solve the second question:
1. Convert the given temperature from Celsius to Kelvin:
T = 667 + 273 = 940 K
2. Convert the given pressure to atm:
P = 11.3 atm
3. Use the given volume of ammonia to find the number of moles of ammonia:
n( NH3 ) = PV / RT
n( NH3 ) = (11.3 atm) * (1.19 X 10^4 L) / (0.0821 L·atm/(mol·K) * 940 K)
4. Since the stoichiometry of the reaction is 1 mole of NH3 yields 1 mole of nitrogen gas (N2), there will be the same number of moles of nitrogen gas as ammonia. Therefore, the number of moles of nitrogen gas is also:
n( N2 ) = n( NH3 )
5. Calculate the volume of nitrogen gas produced using the number of moles of nitrogen gas and the given temperature and pressure:
V( N2 ) = n( N2 ) * (RT / P)
V( N2 ) = n( NH3 ) * (RT / P)
V( N2 ) = (n( NH3 ) * (0.0821 L·atm/(mol·K) * 940 K)) / 11.3 atm