how many liters of ammonia are required to react with 1 mole of oxygen gas at 850 degrees centigrade and 5 atm in order to produce nitrogen monoxide and water vapor at the same conditions
the answer is 15 right
To determine the number of liters of ammonia required, we first need to write the balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) to produce nitrogen monoxide (NO) and water vapor (H2O):
4 NH3 + 5 O2 -> 4 NO + 6 H2O
From the balanced equation, we can see that 4 moles of NH3 react with 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O.
To find the number of liters of ammonia required, we need to consider the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
Let's plug in the given values:
P = 5 atm (pressure)
T = 850 + 273 = 1123 K (temperature)
Now, let's calculate the number of moles of O2. Since we have 1 mole of O2, we can directly use this value in calculations.
n(O2) = 1 mole
To find the number of moles of NH3 required, we can set up a proportion using the stoichiometric coefficients from the balanced equation:
4 moles NH3 = 5 moles O2
x moles NH3 = 1 mole O2 (given)
Using the proportion, we can find:
x = (1 mole O2 * 4 moles NH3) / 5 moles O2
x = 0.8 moles NH3
Now that we have the number of moles of NH3 (0.8 moles), we can use the ideal gas law to calculate the volume of ammonia gas in liters:
V = nRT / P
V = (0.8 moles * 0.0821 L·atm/mol·K * 1123 K) / 5 atm
V ≈ 14.27 liters
Therefore, approximately 14.27 liters of ammonia gas are required to react with 1 mole of oxygen gas at 850 degrees Celsius and 5 atm to produce nitrogen monoxide and water vapor at the same conditions.