Hello, I was hoping that I could get some help with calculating the mass needed to completely react with this question.
In the following chemical reaction between H2 and N2 to produce NH3, what is the mass of N2 needed to completely react with 0.65 g H2 to produce 3.68 g of NH3?
3H2(g) + N2(g) → 2NH3(g)
I know that the chemical equation is already balanced, I am just not really sure where to go from here.
Well, well, well, looks like we've got a chemistry problem on our hands! Don't worry, I've got just the clownish wisdom you need to tackle this one.
To find the mass of N2 needed, we can start by focusing on the balanced equation:
3H2(g) + N2(g) → 2NH3(g)
We know that 0.65 g of H2 reacts to produce 3.68 g of NH3. So, let's start by finding the molar mass of NH3:
NH3 = 1 × (1.01 g/mol) + 3 × (1.01 g/mol) = 17.03 g/mol
Now, we can use stoichiometry to find the molar ratio between H2 and NH3:
3 moles of H2 → 2 moles of NH3
Since we have the mass of H2 (0.65 g), we can convert it to moles:
0.65 g H2 × (1 mol/2.02 g) = 0.322 moles of H2
Now, we can use the molar ratio to find the number of moles of NH3 produced:
0.322 moles of H2 × (2 moles of NH3/3 moles of H2) = 0.215 moles of NH3
Since we know the mass of NH3 (3.68 g) and the molar mass of NH3 (17.03 g/mol), we can find the number of moles of NH3:
3.68 g NH3 × (1 mol/17.03 g) = 0.216 moles of NH3
Now that we know the number of moles of NH3 produced, we can find the number of moles of N2 needed by using the stoichiometric ratio:
2 moles of NH3 → 1 mole of N2
0.215 moles of NH3 × (1 mole of N2/2 moles of NH3) = 0.108 moles of N2
Finally, to find the mass of N2 needed, we can multiply the number of moles of N2 by its molar mass:
0.108 moles of N2 × (28.01 g/mol) = 3.01 g N2
So, to completely react with 0.65 g of H2 and produce 3.68 g of NH3, you would need approximately 3.01 grams of N2. Ta-da!
To find the mass of N2 needed to completely react with the given amount of H2 and produce the given amount of NH3, you can follow these steps:
Step 1: Calculate the number of moles of H2 used:
To do this, use the molar mass of H2, which is 2.02 g/mol.
Number of moles of H2 = mass of H2 / molar mass of H2
Number of moles of H2 = 0.65 g / 2.02 g/mol
Step 2: Use the balanced chemical equation to determine the mole ratio between H2 and N2.
From the balanced equation, the mole ratio of H2 to N2 is 3:1. This means that every 3 moles of H2 react with 1 mole of N2.
Step 3: Calculate the number of moles of N2 needed:
To do this, use the mole ratio between H2 and N2.
Number of moles of N2 = (Number of moles of H2 / mole ratio of H2 to N2)
Number of moles of N2 = (0.65 g / 2.02 g/mol) / (3 mol H2 / 1 mol N2)
Step 4: Calculate the mass of N2 needed:
To do this, use the molar mass of N2, which is 28.02 g/mol.
Mass of N2 needed = Number of moles of N2 * molar mass of N2
By following these steps, you can calculate the mass of N2 needed to completely react with 0.65 g of H2 to produce 3.68 g of NH3.
To calculate the mass of N2 needed to completely react with a given amount of H2, you can use the concept of stoichiometry, which relates the coefficients of the balanced chemical equation to the moles of substances involved in the reaction.
Here are the steps to follow:
Step 1: Convert the mass of H2 to moles.
To do this, you need the molar mass of H2, which is 2.02 g/mol (2 hydrogen atoms, each with a molar mass of 1.01 g/mol).
moles of H2 = mass of H2 / molar mass of H2
moles of H2 = 0.65 g / 2.02 g/mol
Step 2: Use the balanced chemical equation to establish the stoichiometric ratio between H2 and N2.
From the balanced equation, you can see that 3 moles of H2 react with 1 mole of N2.
Step 3: Calculate the moles of N2 required.
Using the ratio from the balanced equation, you can determine how many moles of N2 are needed.
moles of N2 = moles of H2 (from Step 1) / stoichiometric ratio (from the balanced equation)
Step 4: Convert moles of N2 to mass.
To determine the mass of N2 needed, multiply the moles of N2 by its molar mass, which is 28.02 g/mol (2 nitrogen atoms, each with a molar mass of 14.01 g/mol).
mass of N2 = moles of N2 x molar mass of N2
Now let's go through the calculations:
Step 1:
moles of H2 = 0.65 g / 2.02 g/mol ≈ 0.3228 mol H2
Step 2:
The stoichiometric ratio between H2 and N2 is 3:1 (from the balanced equation).
Step 3:
moles of N2 = 0.3228 mol H2 / 3 = 0.1076 mol N2
Step 4:
mass of N2 = 0.1076 mol N2 x 28.02 g/mol ≈ 3.01 g N2
Therefore, the mass of N2 needed to completely react with 0.65 g of H2 to produce 3.68 g of NH3 is approximately 3.01 g.
N = 14 g/mol
H = 1 g/mol
NH3 = 14+3 = 17 g/mol
3.68 g * 14/17 = 3.03 grams of N needed
by the way check grams of H
3.68 *3/17 = .65, just enough hydrogen:)