4 NH3 + 3 O2 --> 2 N2 + 6 H2O
a.) Assume 3.0 mol O2 completely reacts. Calculate the moles required for the other reactant. Calculate the moles that can be produced for each product.
b.) Assume 8.0 mol NH3 completely reacts. Calculate the moles required for the other reactant. Calculate the moles that can be produced for each product.
c.) Assume 1.0 mol N2 is produced. Calculate the moles required for each reactant. Calculate the moles that can be produced for the other product.
d.) Assume 0.40 mol H20 is produced. Calculate the moles required for each reactant. Calculate the moles that can be produced for the other product.
How do you do this?
4 mols of NH3 for every 3 of O2
so 4
4/3 = 8/6
so 6 of O2
also 4/2 = 8/4 so 4 of N2
4/6 = 8/12 so 12 of H2O
I am getting bored :(
4/2 = 2/1 so 2 of NH3
3 of H2O
enough already
To solve these problems, we need to use the stoichiometry of the balanced chemical equation. The coefficients in front of each compound represent the mole ratio between the reactants and products.
Here are the steps to solve each part of the question:
a.) Assume 3.0 mol O2 completely reacts. We want to determine the moles required for the other reactant, NH3, as well as the moles that can be produced for each product.
First, we need to find the mole ratio between NH3 and O2. From the balanced equation, we know that the ratio is 4:3 (4 moles NH3 react with 3 moles O2).
To find the moles of NH3, we can set up a proportion using the given mole of O2:
(3.0 mol O2) x (4 mol NH3 / 3 mol O2) = 4.0 mol NH3
So, 4.0 moles of NH3 are required to react completely when 3.0 moles of O2 are used.
Next, we can use the mole ratio between NH3 and the products N2 and H2O to determine the moles that can be produced.
From the balanced equation, we know that the ratio between NH3 and N2 is 4:2 (4 moles NH3 produce 2 moles N2), and the ratio between NH3 and H2O is 4:6 (4 moles NH3 produce 6 moles H2O).
To find the moles of N2 that can be produced, we can use the mole ratio and the moles of NH3 calculated earlier:
(4.0 mol NH3) x (2 mol N2 / 4 mol NH3) = 2.0 mol N2
Therefore, 2.0 moles of N2 can be produced when 3.0 moles of O2 completely reacts.
Similarly, to find the moles of H2O that can be produced, we can use the mole ratio and the moles of NH3:
(4.0 mol NH3) x (6 mol H2O / 4 mol NH3) = 6.0 mol H2O
So, 6.0 moles of H2O can be produced when 3.0 moles of O2 completely reacts.
b.) Assume 8.0 mol NH3 completely reacts. We want to calculate the moles required for the other reactant, O2, as well as the moles that can be produced for each product.
Following the same steps as in part a, we use the mole ratio between NH3 and O2 (4:3) to calculate the moles of O2:
(8.0 mol NH3) x (3 mol O2 / 4 mol NH3) = 6.0 mol O2
Thus, 6.0 moles of O2 are required to react completely when 8.0 moles of NH3 are used.
The mole ratio between NH3 and N2 (4:2) and NH3 and H2O (4:6) can be used to determine the moles of N2 and H2O that can be produced:
(8.0 mol NH3) x (2 mol N2 / 4 mol NH3) = 4.0 mol N2
(8.0 mol NH3) x (6 mol H2O / 4 mol NH3) = 12.0 mol H2O
Therefore, when 8.0 moles of NH3 completely reacts, 4.0 moles of N2 and 12.0 moles of H2O can be produced.
c.) Assume 1.0 mol N2 is produced. We want to calculate the moles required for each reactant, NH3 and O2, as well as the moles that can be produced for the other product, H2O.
To determine the moles required for NH3, we use the mole ratio between NH3 and N2 (4:2) and the given mole of N2:
(1.0 mol N2) x (4 mol NH3 / 2 mol N2) = 2.0 mol NH3
Thus, 2.0 moles of NH3 are required to produce 1.0 mole of N2.
Next, we can use the mole ratio between NH3 and H2O (4:6) to calculate the moles of H2O that can be produced:
(2.0 mol NH3) x (6 mol H2O / 4 mol NH3) = 3.0 mol H2O
So, when 1.0 mole of N2 is produced, 2.0 moles of NH3 react, and 3.0 moles of H2O can be produced.
To find the moles of O2 required, we use the mole ratio between O2 and N2 (3:2) and the given mole of N2:
(1.0 mol N2) x (3 mol O2 / 2 mol N2) = 1.5 mol O2
Therefore, when 1.0 mole of N2 is produced, 2.0 moles of NH3 react, 3.0 moles of H2O can be produced, and 1.5 moles of O2 are required.
d.) Assume 0.40 mol H2O is produced. We want to calculate the moles required for each reactant, NH3 and O2, as well as the moles that can be produced for the other product, N2.
Using the mole ratio between NH3 and H2O (4:6) and the given mole of H2O:
(0.40 mol H2O) x (4 mol NH3 / 6 mol H2O) = 0.27 mol NH3
Thus, 0.27 moles of NH3 are required to produce 0.40 mole of H2O.
To determine the moles of O2 required, we use the mole ratio between O2 and H2O (3:6) and the given mole of H2O:
(0.40 mol H2O) x (3 mol O2 / 6 mol H2O) = 0.20 mol O2
So, when 0.40 mole of H2O is produced, 0.27 mole of NH3 reacts, and 0.20 mole of O2 is required.
Finally, we can find the moles of N2 that can be produced by using the mole ratio between NH3 and N2 (4:2) and the moles of NH3:
(0.27 mol NH3) x (2 mol N2 / 4 mol NH3) = 0.14 mol N2
Therefore, when 0.40 mole of H2O is produced, 0.27 mole of NH3 reacts, 0.20 mole of O2 is required, and 0.14 mole of N2 can be produced.