4NH3(g) + 5O2(g) -> 4NO(g) + 6H20(g)

what mass of H2O can be produced from 350 liters of NH3?

Hi I put the wrong equation up should read 4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)

Question then is what mass of H2O is produced from 350 liters of NH3?

To calculate the mass of H2O produced from 350 liters of NH3, we need to follow a series of steps:

Step 1: Convert volume to moles.
NH3 is measured in liters, but we need to work with moles for further calculations. To convert the volume of NH3 into moles, we'll use the ideal gas law equation:

PV = nRT

P = pressure (assume 1 atm)
V = volume (350 liters)
n = moles of NH3 (to be calculated)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (usually in Kelvin)

Step 2: Calculate moles of H2O.
According to the balanced equation, 4 moles of NH3 react to produce 6 moles of H2O. Therefore, the molar ratio of NH3 to H2O is 4:6.

Since we know the moles of NH3 from step 1, we can calculate the moles of H2O using a proportion:

(6 moles H2O / 4 moles NH3) = (x moles H2O / moles NH3)

Step 3: Convert moles of H2O to grams.
To convert moles of H2O to grams, we need to multiply the moles by the molar mass of H2O. The molar mass of H2O is approximately 18.02 g/mol.

Step 4: Calculate the final mass of H2O.
Multiply the moles of H2O calculated in step 2 by the molar mass of H2O obtained in step 3 to get the mass of H2O.

Now, let's go through the calculations:

Step 1:
n = (PV) / (RT)
Assuming the temperature is at standard conditions (298 K) and pressure is 1 atm,
n = (1 atm * 350 L) / (0.0821 L·atm/(mol·K) * 298 K)

Step 2:
Moles of H2O = (6 moles H2O / 4 moles NH3) * n moles NH3

Step 3:
Mass of H2O = Moles of H2O * Molar mass of H2O

Step 4:
Calculate the final mass using the equation obtained in step 3.

Please note that the actual numerical answer would depend on the specific conditions given for the temperature and pressure.