A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine (N2H4) and dinitrogen tetraoxide (N2O4), which ignite on contact to form nitrogen gas and water vapor.
2 N2H4 + N2O4 --> 3 N2 + 6 H2O
If 1.50 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are mixed, what is the theoretical yield (in grams) of N2 that can be produced?
is the answer around 444.85g?
To find the theoretical yield of N2 that can be produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed.
Let's calculate the number of moles for each reactant:
1 mole of N2H4 has a molar mass of 32.05 g/mol.
Number of moles of N2H4 = mass / molar mass = 1.50 x 10^2 g / 32.05 g/mol
1 mole of N2O4 has a molar mass of 92.01 g/mol.
Number of moles of N2O4 = mass / molar mass = 2.00 x 10^2 g / 92.01 g/mol
Now, we need to compare the mole ratio of N2H4 to N2O4 with the stoichiometric ratio in the balanced equation:
2 moles of N2H4 : 1 mole of N2O4
Let's calculate the moles of N2 that can be formed for each reactant:
Moles of N2 from N2H4 = (Number of moles of N2H4) x (1 mole of N2 / 2 moles of N2H4)
Moles of N2 from N2O4 = (Number of moles of N2O4) x (3 moles of N2 / 1 mole of N2O4)
Now, we compare the moles of N2 produced from both reactants to determine the limiting reactant:
The smaller value among the moles of N2 from N2H4 and N2O4 represents the limiting reactant.
Finally, we can calculate the theoretical yield of N2:
Theoretical yield (in grams) = (Moles of N2 from the limiting reactant) x (molar mass of N2)
To find the theoretical yield of N2, we need to determine which reactant is limiting and calculate the amount of N2 that can be produced from that reactant.
Step 1: Calculate the molar mass of N2H4 and N2O4.
Molar mass of N2H4 = (2 x atomic mass of N) + (4 x atomic mass of H) = (2 x 14.01 g/mol) + (4 x 1.01 g/mol) = 32.05 g/mol
Molar mass of N2O4 = (2 x atomic mass of N) + (4 x atomic mass of O) = (2 x 14.01 g/mol) + (4 x 16.00 g/mol) = 92.02 g/mol
Step 2: Convert the given masses of N2H4 and N2O4 to moles.
moles of N2H4 = mass of N2H4 / molar mass of N2H4 = 1.50 x 10^2 g / 32.05 g/mol ≈ 4.67 mol
moles of N2O4 = mass of N2O4 / molar mass of N2O4 = 2.00 x 10^2 g / 92.02 g/mol ≈ 2.17 mol
Step 3: Determine the limiting reactant.
The balanced equation indicates that the stoichiometric ratio between N2H4 and N2O4 is 2:1. Therefore, 2 moles of N2H4 react with 1 mole of N2O4. Since we have 4.67 moles of N2H4 and 2.17 moles of N2O4, the N2O4 is limiting as it will be completely consumed.
Step 4: Calculate the moles of N2 that can be produced from N2O4.
From the balanced equation, 1 mole of N2O4 produces 3 moles of N2. Therefore, 2.17 moles of N2O4 will produce (2.17 mol) x (3 mol N2 / 1 mol N2O4) = 6.51 moles of N2.
Step 5: Calculate the theoretical yield of N2 in grams.
The molar mass of N2 is (2 x atomic mass of N) = (2 x 14.01 g/mol) = 28.02 g/mol.
mass of N2 = moles of N2 x molar mass of N2 = 6.51 mol x 28.02 g/mol ≈ 182.6 g
Therefore, the theoretical yield of N2 that can be produced is approximately 182.6 grams.