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N2 + 3H2-> 2NH3

How many liters of NH, will be produced at a temperature of 208.00 degrees celsius and 4.50 torr pressure to consume 26.00 moles of N2?
Round to two decimal places.

1 Answer

2 votes

Answer:

To solve this problem, we can use the ideal gas law and stoichiometry.

First, we need to convert the given temperature to Kelvin:

208.00 degrees Celsius + 273.15 = 481.15 K

Next, we can use the ideal gas law to find the volume of NH3 produced:

PV = nRT

P = 4.50 torr = 0.00592 atm (converting to atm)

V = unknown (what we are trying to find)

n = moles of NH3 produced = 26.00 moles N2 (from stoichiometry)

R = 0.0821 L·atm/K·mol (gas constant)

T = 481.15 K

Solving for V:

V = nRT/P

V = (26.00 mol)(0.0821 L·atm/K·mol)(481.15 K) / (0.00592 atm)

V = 3671.46 L

However, this is the volume of NH3 produced at standard temperature and pressure (STP), which is 0 degrees Celsius (273.15 K) and 1 atm. We need to convert this to the volume at the given temperature and pressure using the combined gas law:

(P1V1/T1) = (P2V2/T2)

P1 = 1 atm (STP pressure)

V1 = 3671.46 L (volume at STP)

T1 = 273.15 K (STP temperature)

P2 = 0.00592 atm (given pressure)

V2 = unknown (what we are trying to find)

T2 = 481.15 K (given temperature)

Solving for V2:

V2 = (P1V1T2) / (P2T1)

V2 = (1 atm)(3671.46 L)(481.15 K) / (0.00592 atm)(273.15 K)

V2 = 315491.48 L or 315491 L (rounded to two decimal places)

Therefore, 315491 L of NH3 will be produced at a temperature of 208.00 degrees Celsius and 4.50 torr pressure to consume 26.00 moles of N2.

User Oana Andone
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