determine how many grams of N2 are produced from the reaction of 8.75 g of H2O2 and 5.43 g of N2H4.
2H2O2+N2H4 4H2O+N2
g N2?
2H2O2 + N2H4 --> 4H2O + N2
This is a limiting reagent (LR) problem. You know that when amounts are given for BOTH reactants.
mols H2O2 = grams/molar mass = approx 0.26 but that's just an estimate. You should redo ALL of these calculations for all are estimates.
mols N2H4 = approx 0.17
Using the coefficients in the balanced equation, convert mols H2O2 to mols of the product (in this case N2). That is
approx 0.26 x (1 mol N2/2 mols H2O2) = approx 0.13
Do the same to convert mols N2H4 to mols N2. You can go through the math but that's approx 0.17 x (1 mol N2/1 mol N2H4) - approx 0.17
In LR problems the answer is ALWAYS the smaller one so N2H4 is the LR and H2O2 is the excerss reagent.
So take the smaller value, convert to grams of the product.
mols N2 x molar mass N2 = grams N2.
Post your work if you get stuck.
ok so far I have gotten
0.257 mol H2O2
0.169 mol N2H4
Im not sure where to go after this
This step follows to convert mols H2O2 to mols N2.
Using the coefficients in the balanced equation, convert mols H2O2 to mols of the product (in this case N2). That is
approx 0.26 x (1 mol N2/2 mols H2O2) = approx 0.13
Then do this step to convert mols N2H4 to mols N2.
Do the same to convert mols N2H4 to mols N2. You can go through the math but that's approx 0.17 x (1 mol N2/1 mol N2H4) - approx 0.17
The smaller number tells you that N2H4 is the limiting reagent. So take that many mols you calculate and
mols N2 x molar mass N2 = grams N2.
To determine how many grams of N2 are produced in the reaction, you can use the concept of stoichiometry. Stoichiometry is a way to calculate the amount of reactants and products involved in a chemical reaction using the balanced equation.
Step 1: Write the balanced chemical equation:
2H2O2 + N2H4 → 4H2O + N2
According to the balanced equation, 2 moles of H2O2 react with 1 mole of N2H4 to produce 1 mole of N2.
Step 2: Convert the given masses of reactants to moles:
Given:
Mass of H2O2 = 8.75 g
Molar mass of H2O2 = 34.02 g/mol (2 H x 1.01 g/mol + 2 O x 16.00 g/mol)
Moles of H2O2 = Mass / Molar mass
Moles of H2O2 = 8.75 g / 34.02 g/mol ≈ 0.257 mol
Given:
Mass of N2H4 = 5.43 g
Molar mass of N2H4 = 32.05 g/mol (2 N x 14.01 g/mol + 4 H x 1.01 g/mol)
Moles of N2H4 = Mass / Molar mass
Moles of N2H4 = 5.43 g / 32.05 g/mol ≈ 0.169 mol
Step 3: Determine the limiting reactant:
To determine the limiting reactant, compare the mole ratio of the reactants in the balanced equation with the actual mole ratio calculated in step 2.
From the balanced equation:
2 moles of H2O2 react with 1 mole of N2H4
Actual mole ratio:
0.257 mol H2O2 : 0.169 mol N2H4
Since the mole ratio of N2H4 is lower than that of H2O2, N2H4 is the limiting reactant.
Step 4: Determine the moles of N2 produced:
From the balanced equation,
1 mole of N2 is produced from 1 mole of N2H4.
Moles of N2 = Moles of N2H4 ≈ 0.169 mol
Step 5: Convert moles of N2 to grams:
To convert moles of N2 to grams, you need to know the molar mass of N2.
Molar mass of N2 = 28.02 g/mol (2 N x 14.01 g/mol)
Mass of N2 = Moles of N2 x Molar mass of N2
Mass of N2 = 0.169 mol x 28.02 g/mol ≈ 4.73 g
Therefore, approximately 4.73 grams of N2 are produced from the given reaction.