Answer :
Final answer:
Using the stoichiometry of the balanced chemical reaction between hydrogen and nitrogen, 127.5 grams of ammonia can be produced from 22.7 grams of hydrogen, 43.96 grams from 36.3 grams of nitrogen, and when both are present, nitrogen is the limiting reactant which limits ammonia production to 43.96 grams.
Explanation:
First, we need to find the molar mass of hydrogen and ammonia to accurately calculate the mass of ammonia that can be produced from the given amounts of hydrogen and nitrogen.
The molar mass of hydrogen (H2) is approximately 2.02 g/mol and for ammonia (NH3) it's around 17.03 g/mol.
A. Ammonia from Hydrogen
We start with 22.7 g of H2:
Moles of H2 = 22.7 g / 2.02 g/mol = 11.24 mol of H2
Using the stoichiometry of the balanced chemical equation (3 H2 : 2 NH3), we can find the moles of ammonia (NH3) that can be produced:
Moles of NH3 = (11.24 mol H2) × (2 mol NH3 / 3 mol H2) = 7.49 mol NH3
Grams of NH3 = 7.49 mol × 17.03 g/mol = 127.5 g NH3
B. Ammonia from Nitrogen
We start with 36.3 g of N2:
The molar mass of nitrogen (N2) is approximately 28.02 g/mol.
Moles of N2 = 36.3 g / 28.02 g/mol = 1.29 mol of N2
Using the stoichiometry of the balanced chemical equation (1 N2 : 2 NH3), we can find the moles of ammonia (NH3) that can be produced:
Moles of NH3 = (1.29 mol N2) × (2 mol NH3 / 1 mol N2) = 2.58 mol NH3
Grams of NH3 = 2.58 mol × 17.03 g/mol = 43.96 g NH3
C. Limiting Reactant Problem
With both 22.7 g of H2 and 36.3 g of nitrogen, the limiting reactant must be found to calculate the maximum amount of NH3 produced.
For H2: Moles of NH3 that can be produced = 7.49 mol (as calculated in part A).
For N2: Moles of NH3 that can be produced = 2.58 mol (as calculated in part B).
The limiting reactant is N2 since it produces less NH3, so the maximum amount of NH3 that can be produced is 43.96 g (as calculated in part B).
Answer:
To determine how many grams of ammonia can be produced in each scenario, we need to use stoichiometry, which is the relationship between the balanced chemical equation and the amounts of reactants and products involved.
A. To find the grams of ammonia produced from 22.7g of hydrogen, we'll use the given balanced chemical equation:
3 H2(g) + N2(g) → 2 NH3(g)
First, we need to calculate the number of moles of hydrogen using its molar mass:
22.7g H2 × (1 mol H2 / 2.02g H2) = 11.2 mol H2
Next, we use the stoichiometric ratio between hydrogen and ammonia from the balanced equation:
11.2 mol H2 × (2 mol NH3 / 3 mol H2) = 7.47 mol NH3
Finally, we convert the moles of ammonia to grams using its molar mass:
7.47 mol NH3 × (17.03g NH3 / 1 mol NH3) = 127g NH3
Therefore, 22.7g of hydrogen can produce 127g of ammonia.
B. To find the grams of ammonia produced from 36.3g of nitrogen, we'll use the given balanced chemical equation:
3 H2(g) + N2(g) → 2 NH3(g)
First, we need to calculate the number of moles of nitrogen using its molar mass:
36.3g N2 × (1 mol N2 / 28.02g N2) = 1.29 mol N2
Next, we use the stoichiometric ratio between nitrogen and ammonia from the balanced equation:
1.29 mol N2 × (2 mol NH3 / 1 mol N2) = 2.58 mol NH3
Finally, we convert the moles of ammonia to grams using its molar mass:
2.58 mol NH3 × (17.03g NH3 / 1 mol NH3) = 43.9g NH3
Therefore, 36.3g of nitrogen can produce 43.9g of ammonia.
C. To solve the limiting reactant problem, we compare the amounts of both reactants and determine which one limits the amount of product formed.
From the previous calculations, we know that 22.7g of hydrogen can produce 127g of ammonia, and 36.3g of nitrogen can produce 43.9g of ammonia.
To find the limiting reactant, we need to calculate the number of moles of each reactant:
22.7g H2 × (1 mol H2 / 2.02g H2) = 11.2 mol H2
36.3g N2 × (1 mol N2 / 28.02g N2) = 1.29 mol N2
From the balanced equation, we see that the stoichiometric ratio between hydrogen and nitrogen is 3:1.
Comparing the moles of hydrogen and nitrogen, we can see that 1.29 mol N2 is less than 11.2 mol H2. Therefore, nitrogen is the limiting reactant.
Using the stoichiometric ratio between nitrogen and ammonia, we find the amount of ammonia produced from 1.29 mol N2:
1.29 mol N2 × (2 mol NH3 / 1 mol N2) × (17.03g NH3 / 1 mol NH3) = 46.7g NH3
Therefore, in this limiting reactant scenario, 22.7g of hydrogen and 36.3g of nitrogen can produce 46.7g of ammonia.