If a reaction occurs between 25 g of Hydrogen and 15 g of Nitrogen, what will be the mass of the products?
To determine the mass of the products, we need to determine the limiting reactant and then use stoichiometry to calculate the mass of the products.
1. Determine the moles of each reactant:
- Moles of hydrogen (H2):
H2 has a molar mass of 2 g/mol, so 25 g H2 is equal to 25 g / 2 g/mol = 12.5 mol H2.
- Moles of nitrogen (N2):
N2 has a molar mass of 28 g/mol, so 15 g N2 is equal to 15 g / 28 g/mol ≈ 0.536 mol N2.
2. Determine the limiting reactant:
The reactant that produces the fewer moles of product is the limiting reactant:
- The balanced chemical reaction for the formation of ammonia (NH3) from hydrogen and nitrogen is:
N2 + 3H2 → 2NH3.
From the balanced equation, we see that for every 1 mol of N2, three moles of H2 are required. So, if 0.536 mol N2 requires (3 x 0.536) = 1.608 mol H2, and we have 12.5 mol H2, then the limiting reactant is N2.
3. Calculate the mass of the products:
Using the mole ratio from the balanced equation, we can determine the moles of NH3 produced:
From the balanced equation, 1 mole of N2 produces 2 moles of NH3.
Thus, 0.536 mol N2 produces (2 x 0.536) = 1.072 mol NH3.
Now we can calculate the mass of NH3 produced:
- The molar mass of NH3 is 17 g/mol.
So, 1.072 mol NH3 is equal to 1.072 mol x 17 g/mol = 18.224 g NH3.
Therefore, the mass of the products produced from the given reactants is approximately 18.224 grams.