Nitrogen gas reacts with hydrogen gas to produce ammonia via the following reaction:
N2 (g) + 3H2(g) →2N3 (g)
• Part A
If you have 3.32 g of H2, how many grams of NH3 can be produced?
Express your answer with the appropriate units.
To solve this problem, we need to use stoichiometry and the given balanced equation.
From the balanced equation, we can see that the ratio between H2 and NH3 is 3:2.
First, we need to convert the given mass of H2 to moles:
Molar mass of H2 = 2 g/mol
Number of moles of H2 = 3.32 g / 2 g/mol = 1.66 mol
Next, we can use the mole ratio to find the number of moles of NH3 produced:
1 mol H2 : 2 mol NH3
1.66 mol H2 : x mol NH3
Using the ratio, we can set up a proportion:
1.66 mol H2 / (1 mol H2 : 2 mol NH3) = x mol NH3
Simplifying the proportion:
x = 1.66 mol H2 * (2 mol NH3 / 1 mol H2) = 3.32 mol NH3
Finally, we need to convert the moles of NH3 to grams:
Molar mass of NH3 = 17 g/mol
Mass of NH3 = 3.32 mol NH3 * 17 g/mol = 56.44 g
Therefore, 3.32 g of H2 can produce 56.44 g of NH3.
To find out how many grams of NH3 can be produced, we need to use the stoichiometry of the reaction. From the balanced equation, we can see that the mole ratio between N2 and NH3 is 1:2. This means that for every mole of N2 reacted, we get 2 moles of NH3.
First, we need to convert the given mass of H2 to moles. To do this, we can use the molar mass of H2, which is 2.02 g/mol.
Mass of H2 = 3.32 g
Molar mass of H2 = 2.02 g/mol
Number of moles of H2 = Mass of H2 / Molar mass of H2
= 3.32 g / 2.02 g/mol
Next, we use the mole ratio to determine the number of moles of NH3 that can be produced. Since the mole ratio is 3:2 between H2 and NH3, we multiply the number of moles of H2 by the mole ratio:
Number of moles of NH3 = Number of moles of H2 * (2 moles of NH3 / 3 moles of H2)
= (3.32 g / 2.02 g/mol) * (2 mol NH3 / 3 mol H2)
Finally, we can convert the number of moles of NH3 to grams by multiplying with the molar mass of NH3, which is 17.03 g/mol.
Mass of NH3 = Number of moles of NH3 * Molar mass of NH3
= (3.32 g / 2.02 g/mol) * (2 mol NH3 / 3 mol H2) * 17.03 g/mol
Simplifying the expression:
Mass of NH3 = (3.32 * 2 * 17.03) / (2.02 * 3) g
= 11.08 g
Therefore, 11.08 grams of NH3 can be produced when 3.32 grams of H2 react with N2.
To determine the number of grams of NH3 that can be produced, you need to use the stoichiometry of the reaction and the given amount of H2.
Here's how you can solve it step by step:
Step 1: Calculate the number of moles of H2:
To do this, you need to know the molar mass of H2. The molar mass of H2 is 2.016 g/mol.
Number of moles of H2 = mass of H2 / molar mass of H2
= 3.32 g / 2.016 g/mol
≈ 1.646 mol
Step 2: Use stoichiometry to find the number of moles of NH3:
From the balanced equation, you can see that the stoichiometric ratio between H2 and NH3 is 3:2. This means that for every 3 moles of H2, 2 moles of NH3 are produced.
Number of moles of NH3 = (Number of moles of H2) × (2 moles of NH3 / 3 moles of H2)
≈ 1.646 mol × (2/3)
≈ 1.097 mol
Step 3: Convert moles of NH3 to grams:
To do this, you need to know the molar mass of NH3. The molar mass of NH3 is 17.03 g/mol.
Mass of NH3 = (Number of moles of NH3) × (molar mass of NH3)
= 1.097 mol × 17.03 g/mol
≈ 18.66 g
Therefore, approximately 18.66 grams of NH3 can be produced when you have 3.32 grams of H2.