How do you do this?
4 NH3 + 3 O2 --> 2 N2 + 6 H2O
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.
Come on, it is proportion.
If 1mol N2 is produced, then 2 mol NH3 and 1.5molO2 3 mol water is produced.
To solve these problems, we need to use the balanced chemical equation and the mole ratios to calculate the moles of reactants required and the moles of products produced.
c.) Assume 1.0 mol N2 is produced:
According to the balanced equation, 4 moles of NH3 react with 3 moles of O2 to produce 2 moles of N2. So, we can set up the following ratios:
- NH3 : O2 : N2 = 4 : 3 : 2
To calculate the moles required for each reactant, we set up the following proportion:
(1.0 mol N2 / 2 mol N2) = (x mol NH3 / 4 mol NH3)
Solving for x, we get:
x = (1.0 mol N2 * 4 mol NH3) / 2 mol N2 = 2.0 mol NH3
Hence, 2.0 moles of NH3 are required to produce 1.0 mole of N2.
To calculate the moles that can be produced for the other product, we set up the following proportion:
(1.0 mol N2 / 2 mol N2) = (y mol H2O / 6 mol H2O)
Solving for y, we get:
y = (1.0 mol N2 * 6 mol H2O) / 2 mol N2 = 3.0 mol H2O
Hence, 3.0 moles of H2O can be produced when 1.0 mole of N2 is produced.
d.) Assume 0.40 mol H2O is produced:
Using the same approach, we set up the following proportions:
- NH3 : O2 : H2O = 4 : 3 : 6
To calculate the moles required for each reactant, we set up the following proportion:
(0.40 mol H2O / 6 mol H2O) = (x mol NH3 / 4 mol NH3)
Solving for x, we get:
x = (0.40 mol H2O * 4 mol NH3) / 6 mol H2O = 0.27 mol NH3
Hence, 0.27 moles of NH3 are required to produce 0.40 mole of H2O.
To calculate the moles that can be produced for the other product, we set up the following proportion:
(0.40 mol H2O / 6 mol H2O) = (y mol N2 / 2 mol N2)
Solving for y, we get:
y = (0.40 mol H2O * 2 mol N2) / 6 mol H2O = 0.13 mol N2
Hence, 0.13 moles of N2 can be produced when 0.40 mole of H2O is produced.