How much ammonia is produced if 1.80 m3 of hydrogen reacts with 0.5m3 of nitrogen? Also state which reagent is in excess and by how much?
Question is related to the Haber process. Please can anyone answer this for me by 02 July 2010.
Thanx
First write the balanced reaction.
3H2 + N2 => 2 NH3
If the reaction were stoichiometric, there would be three times as many moles of H2 as N2. That also means three times the volume.
In your case, there are more than 3 times as many moles of H2 as N2.
It should be obvious which reactant is in excess, and by how much.
1.0 m^3 of NH3 will be formed, when the products are brought to the pressure and temperature of the reactants. If you know P and T, you can also state the number of moles formed. There are 22.4 liters per mole at STP.
Thank you. :)
To determine the amount of ammonia produced and identify the excess reagent, we can use the balanced equation for the Haber process:
N2 + 3H2 -> 2NH3
1. Calculate the number of moles of hydrogen (H2) and nitrogen (N2) using the Ideal Gas Law equation: PV = nRT
First, convert the given volumes to the corresponding number of moles using the ideal gas equation:
n(H2) = (P)(V) / (R)(T)
= (P)(1.80) / (R)(T)
n(N2) = (P)(V) / (R)(T)
= (P)(0.50) / (R)(T)
Where P is the pressure, R is the ideal gas constant, and T is the temperature.
2. Compare the ratio of moles of H2 to N2 in the balanced equation to determine the limiting reagent.
From the balanced equation, we can see that 3 moles of H2 react with 1 mole of N2 to produce 2 moles of NH3. Therefore, the ratio of H2 to N2 is 3:1.
If the ratio of moles of H2 to N2 calculated in step 1 is also 3:1, both reactants are in the stoichiometric ratio, and neither is in excess.
If the ratio of moles of H2 to N2 calculated in step 1 is less than 3:1, then H2 is the limiting reagent, and N2 is in excess.
If the ratio of moles of H2 to N2 calculated in step 1 is greater than 3:1, then N2 is the limiting reagent, and H2 is in excess.
3. Calculate the mole ratio of NH3 produced based on the limiting reagent.
Using the balanced equation, we know that 3 moles of H2 produce 2 moles of NH3. So, if H2 is the limiting reagent, the mole ratio of H2 to NH3 is 3:2.
If N2 is the limiting reagent, the mole ratio of N2 to NH3 is 1:2.
4. Calculate the moles of NH3 produced based on the limiting reagent.
If H2 is the limiting reagent:
n(NH3) = (2/3)(n(H2))
If N2 is the limiting reagent:
n(NH3) = (2)(n(N2))
5. Convert moles of NH3 to the volume using the Ideal Gas Law.
V(NH3) = (n(NH3))(R)(T) / P
Now, you have obtained the equations and steps needed to calculate the amount of ammonia produced and determine the excess reagent. You can continue with the calculations by plugging in the appropriate values for the pressure, temperature, and gas constant. Since you have mentioned that this question needs to be answered by July 2, 2010, I assume that the deadline has already passed.