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.
With all equations.
Thanx
Yes and no. We help do homework but we don't do it for you. Here is the way you go about solving the question. Most of the limiting reagent problems are done this way.
1. Write and balance the equation.
N2 + 3H2 ==> 2NH3
2. When the equation is all gases, one may take a shortcut and NOT convert to moles.
a. Using the coefficients in the balanced equation, convert volume of N2 to volume of NH3.
1.8 m^3 H2 x (2 moles NH3/3 moles H2) = 1.8 x (2/3) = ?? m^3 NH3 from H2.
2b. Do the same for 0.5 m^3 N2.
2c. It is likely that the volume of NH3 produced according to 2a and 2b will not be the same which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
2d. Now that you know how much NH3 is produced and you have determined the limiting reagent, you turn attention to how much of the "other" reagent was used since you know ALL of the limiting reagent was used.
Same procedure. Use the coefficients to convert volume of the limiting reagent to volume of the "other" reagent. That will tell you how much was used and the difference between the initial value and the amount used will tell you how much is left. Post your work if you get stuck.
To find out how much ammonia is produced in the reaction between hydrogen and nitrogen, we need to first write the balanced equation for the reaction.
The reaction between hydrogen and nitrogen to produce ammonia can be represented by the equation:
N2 + 3H2 ➝ 2NH3
From the balanced equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
To determine the amount of ammonia produced, we need to calculate the moles of both nitrogen and hydrogen used in the reaction.
Given:
Volume of hydrogen = 1.80 m3
Volume of nitrogen = 0.5 m3
To calculate moles, we need to use the ideal gas equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature.
Since the volumes of the gases are given, we can use the equation V = nRT/P to solve for the number of moles of each gas.
First, let's calculate the moles of hydrogen (H2):
Using the ideal gas equation:
n(H2) = (V(H2) x P) / (R x T)
Assuming the temperature is constant at standard temperature and pressure (STP), which is 0°C (273 K) and 1 atm, respectively, and substituting the given values:
n(H2) = (1.80 m3 x 1 atm) / (0.0821 L·atm/mol·K x 273 K)
n(H2) ≈ 78.27 moles
Next, let's calculate the moles of nitrogen (N2):
Using the ideal gas equation:
n(N2) = (V(N2) x P) / (R x T)
Substituting the given values:
n(N2) = (0.5 m3 x 1 atm) / (0.0821 L·atm/mol·K x 273 K)
n(N2) ≈ 6.14 moles
According to the balanced equation, the reaction consumes 3 moles of hydrogen for every 1 mole of nitrogen. We can compare the moles of hydrogen and nitrogen to determine which reactant is in excess.
From the calculations:
n(H2) = 78.27 moles
n(N2) = 6.14 moles
To determine the limiting reagent, we need to calculate the moles of each reactant needed to react in a 3:1 ratio.
For hydrogen:
n(H2 needed) = 6.14 moles x (3 moles H2/1 mole N2)
n(H2 needed) ≈ 18.42 moles
Comparing the moles of hydrogen needed (18.42 moles) to the actual moles of hydrogen provided (78.27 moles), we can see that hydrogen is in excess.
To calculate the moles of ammonia produced, we use the ratio from the balanced equation:
For every 3 moles of hydrogen, 2 moles of ammonia are produced.
n(NH3) = n(H2) x (2 moles NH3/3 moles H2)
n(NH3) ≈ 52.18 moles
Therefore, approximately 52.18 moles of ammonia will be produced.
In conclusion:
- The amount of ammonia produced is approximately 52.18 moles.
- Hydrogen is the excess reagent, with an excess amount of approximately 78.27 - 18.42 = 59.85 moles.
Please note that the calculations assume ideal gas behavior and the reactants are at STP.