How many ammonia liters of ammonia gas are produced when 85 grams of liquid nitrogen completely react?

N2 + 3H2 = 2NH3

One has to know pressure and temperature to find volumes of gas given moles or mass. I am getting somewhat annoyed that this is being reposted.

The only way to report the yield with the data given is in 'general' terms of number of molar Volumes (nVm). BP is correct, you will need Temp and Pressure values to calculate specific volumes.

N2 + 3H2 => 2NH3
85gms N2 = 3.04mole N2 => 2(3.04mole NH3) = 6.08Vm NH3. Vm is Temperature-Pressure dependent, but given (P,V) ... from the Ideal Gas Law Vm = (nRT/P)/(moles NH3) => Specific molar Volume.

To find out how many liters of ammonia gas are produced when 85 grams of liquid nitrogen completely react, we need to follow these steps:

1. Convert the mass of liquid nitrogen (in grams) to moles. We can do this by using the molar mass of nitrogen (N2), which is approximately 28 grams/mol.

Number of moles of N2 = Mass of N2 / Molar mass of N2
= 85 grams / 28 grams/mol

2. Determine the stoichiometric ratio between nitrogen (N2) and ammonia (NH3) using the balanced equation:

N2 + 3H2 = 2NH3

From the equation, we see that 1 mole of N2 reacts to produce 2 moles of NH3.

3. Use the stoichiometric ratio to find the number of moles of NH3 produced. Since the ratio is 1:2 (N2:NH3), we can multiply the number of moles of N2 by 2:

Number of moles of NH3 produced = Number of moles of N2 * 2

4. Convert the number of moles of NH3 to volume (in liters) using the ideal gas law. The ideal gas law equation is:

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.

Since we are given the number of moles (n), we can rearrange the equation to solve for V:

V = nRT / P

The standard temperature and pressure (STP) conditions are usually used for these kinds of calculations, where the pressure (P) is 1 atm and the temperature (T) is 273.15 K.

Note: The ideal gas constant (R) is typically 0.0821 L·atm/(mol·K).

Now, let's calculate the answer:

1. Moles of N2 = 85 g / 28 g/mol = 3.04 mol

2. Moles of NH3 produced = 3.04 mol * 2 = 6.08 mol

3. Volume of NH3 produced at STP:
V = (6.08 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm) = 13.6 L

Therefore, approximately 13.6 liters of ammonia gas are produced when 85 grams of liquid nitrogen completely react.