"According to the equation, for 1 molecule of nitrogen (N2) gas and 3 molecules of hydrogen (H2) gas, 2 molecules of ammonia gas (NH3) are produced. So the mole ratio of Nitrogen to Ammonia is 1:2.
Assuming both Nitrogen gas and Ammonia gas are close approximations to an ideal gas, a mole of each occupies the same volume. So the volume ratio is the same as the mole ratio."
(Vammonia) / 4.74L = 2 / 1
So it's twice the volume.
Hope I helped you...
What will be the volume of ammonia formed from 2L of nitrogen and 2L of hydrogen
2 answers
I disagree with Nerd. This is a limiting reagent (LR) problem.
N2 + 3H2 ==> 2NH3
a. Convert 2 L N2 to mols NH3 produced if we had all of the H2 needed. That is 2 L N2 x (2 mols NH3/1 mol N2) = 2 x 2 = 4 L NH3
b. Convert 2 L H2 to mols NH3 produced if we had all of the N2 needed. That is 2 L H2 x (2 mols NH3/3 mols H2) = 2 x 2/3 = 4/3 or 1-1/3 L NH3.
c. The two answers for L NH3 produced don't agree; in LR problems the SMALLER amount is ALWAYS the correct value to use and the reactant responsible for that value is the LR. So you will produce 1-1/3 L NH3, use up all of the H2 and you will have some N2 remaining after the reaction is complete. The amount of N2 remaining un-reacted can be calculated if you wanted to do that.
N2 + 3H2 ==> 2NH3
a. Convert 2 L N2 to mols NH3 produced if we had all of the H2 needed. That is 2 L N2 x (2 mols NH3/1 mol N2) = 2 x 2 = 4 L NH3
b. Convert 2 L H2 to mols NH3 produced if we had all of the N2 needed. That is 2 L H2 x (2 mols NH3/3 mols H2) = 2 x 2/3 = 4/3 or 1-1/3 L NH3.
c. The two answers for L NH3 produced don't agree; in LR problems the SMALLER amount is ALWAYS the correct value to use and the reactant responsible for that value is the LR. So you will produce 1-1/3 L NH3, use up all of the H2 and you will have some N2 remaining after the reaction is complete. The amount of N2 remaining un-reacted can be calculated if you wanted to do that.
1 answer