Nitrogen & hydrogen goes react to form ammonia gas as follows:
N2(g)+3H2(g)-->2NH3(g)
At a certain temperature & pressure, 1.2L of N2 reacts with 3.6L of H2. If all the N2 and H2 are consumed, what volume of NH3, at the same temperature & pressure will be produced?
I tried working it out with PV=nRT to find the moles of each reactant and then combined them to use PV=nRT again to find the volume of ammonia, but I'm not sure if that's the correct answer or procedure.
An explanation would be great as to how I'm supposed to go about this problem and why. Thanks
Ah, the wonderful world of chemistry equations! Let's see if we can figure this out together, shall we?
The balanced equation you provided is indeed correct:
N2(g) + 3H2(g) --> 2NH3(g)
Now, to find the volume of NH3 gas produced, we need to follow a few steps.
Step 1: Convert the given volumes of N2 and H2 to moles using the ideal gas law equation, PV = nRT.
Given:
Volume of N2 (V1) = 1.2 L
Volume of H2 (V2) = 3.6 L
Step 2: Calculate the moles of N2 and H2.
Using the ideal gas law equation, rearranged to solve for n:
n = PV/RT
Assuming the temperature, pressure, and gas constant (R) remain constant, we can calculate the moles of each gas.
Step 3: Determine the limiting reactant.
Compare the mole ratio of N2 to H2 in the balanced equation (1:3) to find the limiting reactant. Whichever reactant is in excess will not be completely consumed.
Step 4: Use the mole ratio to determine the moles of NH3 produced.
From the balanced equation, we can see that every 1 mole of N2 reacts to produce 2 moles of NH3.
Step 5: Convert moles of NH3 to volume using the ideal gas law.
Using the balanced equation, we can determine the number of moles of NH3 produced. Then, using the ideal gas law equation, we can find the volume of NH3 gas produced.
Remember, at the same temperature & pressure, the volume of the gases is directly proportional to the number of moles.
So, to sum it up, follow these steps:
1. Convert volumes of N2 and H2 to moles.
2. Compare the mole ratio to determine the limiting reactant.
3. Calculate the moles of NH3 produced using the limiting reactant.
4. Convert moles of NH3 to volume using the ideal gas law.
Now, let's crunch some numbers and find out how much ammonia gas will be produced!
To solve this problem, you can use the ideal gas law (PV = nRT) to find the number of moles of each gas involved.
First, let's start by finding the number of moles of N2 and H2:
For N2:
PV = nRT
n = PV/RT
= (1 atm)(1.2 L) / (0.0821 atm·L/mol·K)(given temperature in K)
For H2:
PV = nRT
n = PV/RT
= (1 atm)(3.6 L) / (0.0821 atm·L/mol·K)(given temperature in K)
Now that you have the moles of N2 and H2, we can look at the stoichiometry of the reaction: 1 mol N2 reacts with 3 mol H2 to form 2 mol NH3.
From the previous calculations, you can determine the number of moles of NH3 formed.
Since the ratio of moles of N2 to NH3 is 1:2, the number of moles of NH3 will be twice the number of moles of N2 consumed.
Afterward, you can use the ideal gas law again to determine the volume of NH3 produced.
Now, you can follow these steps to find the volume of NH3 produced at the same temperature and pressure:
1. Calculate the moles of N2 and H2 using the ideal gas law.
2. Determine the moles of NH3 formed based on the stoichiometry of the reaction.
3. Use the ideal gas law again to find the volume of NH3 produced by plugging in the new number of moles into the equation PV = nRT.
Make sure to use the same temperature and pressure throughout the calculations.
I hope this explanation clarifies the procedure for you.
To solve this problem, you can use the ideal gas law equation, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. However, in this case, you do not need to know the actual values for pressure, temperature, or the gas constant, since they remain constant throughout the problem.
The first step is to find the number of moles of N2 and H2 using the given volumes. To do this, you need to know the relationship between volume and moles of a gas at constant temperature and pressure. According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain an equal number of molecules.
From the balanced chemical equation, you can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the ratio of moles of N2 to H2 to NH3 is 1:3:2.
Given that the volume of N2 is 1.2L, you can calculate the number of moles of N2 using the volume-to-mole ratio:
n(N2) = V(N2) / V(molar)
where V(molar) is the molar volume.
Similarly, calculate the number of moles of H2 using the volume-to-mole ratio:
n(H2) = V(H2) / V(molar)
Once you have the number of moles of N2 and H2, you need to determine which reactant is limiting. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, compare the moles of N2 and H2 using the stoichiometry of the balanced equation:
n(NH3) = (n(N2) / mole ratio(N2)) * mole ratio(NH3)
and
n(NH3) = (n(H2) / mole ratio(H2)) * mole ratio(NH3)
The smaller value obtained between the two calculations will be the number of moles of NH3 produced.
Finally, convert the moles of NH3 to volume using the volume-to-mole ratio:
V(NH3) = n(NH3) * V(molar)
Therefore, following this procedure, you can find the volume of NH3 produced when all the N2 and H2 are consumed at the given temperature and pressure.
Thanks so much, that really helped me understand it.
There is a long way and a shortcut way. First the long way which will work on all problems.
This is a simple (as opposed to limiting reagent stoichiometry problem) stoichiometry problem. (USUALLY we know it is a limiting reagent problem when amounts for BOTH reactants are given; however, in this one the problem states that BOTH reactants are consumed entirely.)
N2 + 3H2 ==> 2NH3
First you do PV = nRT to obtain mols N2 and H2. You started off on the right foot although you needed only one of them.
Next, using the coefficients in the balanced equation, use either mols N2 or mols H2 to convert to mols NH3.
Then you can use PV = nRT and convert mols NH3 to volume.
The short way. When all are gases involved we need not convert to mols; i.e., we can use volume directly.
Therefore, convert 1.2L N2 to L NH3.
1.2L N2 x (2 mols NH3/1 mol N2) = 2.4 L NH3.
Or you could use L H2 the same way.
3.6 L H2 x (2 mols NH3/3 mol H2) = 2.4 L NH3.