3H2+N2=2NH3
if 2.70 g of H2 reacts with jugful of N ,how many grams of NH3 is formed?
I assume a jugful means an excess of N2.
mol H2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols H2 to mols NH3.
Now convert mols NH3 to grams. g = mols NH3 x molar mass NH3.
i am sorry i don't understand!
molar mass for what? just for H2 or for 3h2 and N2 also?
To determine the amount of NH3 formed, we need to use the concept of stoichiometry. The balanced chemical equation you provided is:
3H2 + N2 -> 2NH3
From the equation, we can see that the ratio between H2 and NH3 is 3:2. This means that for every 3 moles of H2, we produce 2 moles of NH3.
To solve the problem, we will follow these steps:
Step 1: Convert the mass of H2 to moles.
Step 2: Use the mole ratio to find the moles of NH3.
Step 3: Convert the moles of NH3 to grams.
Step 1: Converting the mass of H2 to moles.
The molar mass of H2 is 2 g/mol.
Moles of H2 = Mass of H2 / Molar mass of H2
Moles of H2 = 2.70 g / 2 g/mol
Moles of H2 = 1.35 mol
Step 2: Using the mole ratio to find the moles of NH3.
The ratio between H2 and NH3 is 3:2.
Moles of NH3 = (Moles of H2 / 3) * 2
Moles of NH3 = (1.35 mol / 3) * 2
Moles of NH3 = 0.90 mol
Step 3: Converting the moles of NH3 to grams.
The molar mass of NH3 is 17 g/mol.
Mass of NH3 = Moles of NH3 * Molar mass of NH3
Mass of NH3 = 0.90 mol * 17 g/mol
Mass of NH3 = 15.3 g
Therefore, when 2.70 g of H2 reacts with N, approximately 15.3 g of NH3 is formed.
To solve this problem, we need to use stoichiometry and the concept of mole ratios.
First, let's calculate the number of moles of H2. The molar mass of H2 is 2.02 g/mol.
Given that we have 2.70 g of H2, we can calculate the number of moles using the formula:
number of moles = mass / molar mass
number of moles of H2 = 2.70 g / 2.02 g/mol ≈ 1.3376 mol of H2
Next, we need to determine the mole ratio between H2 and NH3 from the balanced chemical equation.
From the equation: 3H2 + N2 → 2NH3, we can see that the mole ratio is 3:2 for H2:NH3.
Now, we can calculate the number of moles of NH3 formed using the mole ratio:
number of moles of NH3 = (number of moles of H2) × (moles of NH3 / moles of H2)
number of moles of NH3 = 1.3376 mol × (2 mol NH3 / 3 mol H2) ≈ 0.8917 mol NH3
Finally, we can find the mass of NH3 formed using the molar mass of NH3, which is 17.03 g/mol.
mass of NH3 = number of moles of NH3 × molar mass
mass of NH3 = 0.8917 mol × 17.03 g/mol ≈ 15.18 g
Therefore, approximately 15.18 grams of NH3 is formed when 2.70 grams of H2 reacts with a jugful of N.