How many grams of NH3 can be produced
from the reaction of 17.8 moles of H2 and a sufficient supply of N2?
N2 + 3H2 ---> 2NH3
202 grams
To determine the number of grams of NH3 that can be produced, we follow these steps:
Step 1: Determine the molar ratio between N2 and NH3.
From the balanced equation, we can see that 1 mole of N2 reacts to produce 2 moles of NH3.
Step 2: Convert the moles of N2 to moles of NH3.
We have 17.8 moles of H2, and since there is a sufficient supply of N2, the moles of N2 will be in the same ratio as the moles of H2. Therefore, we have 17.8 moles of N2.
Using the molar ratio from step 1, we can calculate the moles of NH3 that can be produced:
17.8 moles N2 * (2 moles NH3 / 1 mole N2) = 35.6 moles NH3
Step 3: Convert the moles of NH3 to grams.
To convert moles to grams, we need to multiply by the molar mass of NH3.
The molar mass of NH3 is calculated as follows:
Nitrogen (N) has a molar mass of 14.01 g/mol.
Hydrogen (H) has a molar mass of 1.01 g/mol.
NH3 has three hydrogens, so the molar mass of NH3 is:
1 mol N + 3 mol H = 14.01 g/mol + (3 * 1.01 g/mol) = 17.04 g/mol
Now, we can calculate the mass of NH3:
35.6 moles NH3 * 17.04 g/mol = 605.184 g
So, approximately 605.18 grams of NH3 can be produced from the reaction.
Use the coefficients to convert mols H2 to mols NH3.
Convert mols NH3 to grams. g = mol x molar mass.