A mixture of hydrogen peroxide, H2O2, and hydrazine, N2H4, can be used
as a rocket propellant. The reaction is:
7 H2O2(g) + N2H4(l) ® 2 HNO3(aq) + 8 H2O(l)
a) How many moles of H2O2 react with 0.477 mol N2H4? [1] ___________
b) How many grams of HNO3 can be produced in a reaction of 67.7 g
H2O2 with excess N2H4? [3]
c) How many grams of HNO3 can be produced in a reaction of 67.7 g
H2O2 with 10.00 mL of N2H4 ( d= 1.006 g/mL) if the yield of the reaction
is 76.5 %? [4]
A 3.34
b 35.95
c 27.4 g
A is correct.
For B I obtained 35.8.
For C you have assumed (or calculated) that H2O2 is the limiting reagent. If that were correct, then your answer of 27.4 is ok; however, N2H4 is the limiting reagent and your answer is not correct. I calculate something like 15 g HNO3 but check my work. I just did a quickie calculation.
To determine the answers to the given questions, we'll use stoichiometry, which is a method of calculating the relationships between reactants and products in a chemical reaction.
a) How many moles of H2O2 react with 0.477 mol N2H4?
In the balanced chemical equation, we see that the stoichiometric ratio between H2O2 and N2H4 is 7:1.
So, we can set up a proportion to find the number of moles of H2O2:
7 mol H2O2 / 1 mol N2H4 = x mol H2O2 / 0.477 mol N2H4
Cross-multiplying and solving for x, we get:
x = (7 mol H2O2 / 1 mol N2H4) * 0.477 mol N2H4
x = 3.34 mol H2O2
Therefore, 3.34 moles of H2O2 react with 0.477 mol N2H4.
b) How many grams of HNO3 can be produced in a reaction of 67.7 g H2O2 with excess N2H4?
First, we need to convert the mass of H2O2 to moles. The molar mass of H2O2 is 34.02 g/mol.
Number of moles of H2O2 = 67.7 g / 34.02 g/mol
Number of moles of H2O2 = 1.99 mol H2O2
From the balanced chemical equation, we can see that the stoichiometric ratio between H2O2 and HNO3 is 7:2.
So, we can set up a proportion to find the number of moles of HNO3:
7 mol H2O2 / 2 mol HNO3 = 1.99 mol H2O2 / x mol HNO3
Cross-multiplying and solving for x, we get:
x = (2 mol HNO3 / 7 mol H2O2) * 1.99 mol H2O2
x = 0.57 mol HNO3
Finally, we can convert the moles of HNO3 to grams using the molar mass of HNO3:
Mass of HNO3 = 0.57 mol HNO3 * (63.01 g/mol)
Mass of HNO3 = 35.95 g HNO3
Therefore, 35.95 grams of HNO3 can be produced in a reaction of 67.7 g H2O2 with excess N2H4.
c) How many grams of HNO3 can be produced in a reaction of 67.7 g H2O2 with 10.00 mL of N2H4 (d = 1.006 g/mL) if the yield of the reaction is 76.5%?
First, we need to convert the volume of N2H4 from mL to grams.
Mass of N2H4 = volume of N2H4 * density of N2H4
Mass of N2H4 = 10.00 mL * 1.006 g/mL
Mass of N2H4 = 10.06 g N2H4
Now, we can follow the same steps as in part b to find the moles of HNO3 produced and then convert to grams.
Number of moles of H2O2 = 67.7 g / 34.02 g/mol
Number of moles of H2O2 = 1.99 mol H2O2
Number of moles of HNO3 = (2 mol HNO3 / 7 mol H2O2) * 1.99 mol H2O2
Number of moles of HNO3 = 0.57 mol HNO3
Considering a yield of 76.5%, we can calculate the actual amount of HNO3 produced:
Actual mass of HNO3 = 0.57 mol HNO3 * (63.01 g/mol) * 0.765
Actual mass of HNO3 = 27.4 g HNO3
Therefore, considering a yield of 76.5%, 27.4 grams of HNO3 can be produced in a reaction of 67.7 g H2O2 with 10.00 mL of N2H4 (d = 1.006 g/mL).
To answer these questions, we need to use stoichiometry and molar ratios. Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction.
To determine the number of moles of H2O2 that react with 0.477 mol N2H4 (question a):
Step 1: Determine the mole ratio between H2O2 and N2H4 from the balanced equation.
From the balanced equation, we can see that 7 moles of H2O2 react with 1 mole of N2H4.
Step 2: Apply the mole ratio to calculate the number of moles of H2O2.
Multiply the mole ratio by the given number of moles of N2H4.
Mole ratio: 7 moles H2O2 / 1 mole N2H4
0.477 mol N2H4 * (7 mol H2O2 / 1 mol N2H4) = 3.34 moles H2O2
So, the answer to question a is 3.34 moles of H2O2.
To determine the grams of HNO3 produced with 67.7 g H2O2 (question b):
Step 1: Determine the mole ratio between H2O2 and HNO3 from the balanced equation.
From the balanced equation, we can see that 7 moles of H2O2 produce 2 moles of HNO3.
Step 2: Convert grams of H2O2 to moles.
Given: 67.7 g H2O2
Molar mass of H2O2: 34.02 g/mol
67.7 g H2O2 * (1 mol H2O2 / 34.02 g H2O2) = 1.99 mol H2O2
Step 3: Apply the mole ratio to calculate the moles of HNO3.
Mole ratio: 7 moles H2O2 / 2 moles HNO3
1.99 mol H2O2 * (2 mol HNO3 / 7 mol H2O2) = 0.57 mol HNO3
Step 4: Convert moles of HNO3 to grams.
Molar mass of HNO3: 63.02 g/mol
0.57 mol HNO3 * (63.02 g HNO3 / 1 mol HNO3) = 35.95 g HNO3
So, the answer to question b is 35.95 grams of HNO3.
To determine the grams of HNO3 produced with 67.7 g H2O2 and 10.00 mL N2H4 (question c):
Step 1: Determine the number of moles of N2H4 using its density.
Given: 10.00 mL N2H4, density = 1.006 g/mL
10.00 mL N2H4 * (1.006 g N2H4 / 1 mL N2H4) = 10.06 g N2H4
Step 2: Convert grams of N2H4 to moles.
Molar mass of N2H4: 32.05 g/mol
10.06 g N2H4 * (1 mol N2H4 / 32.05 g N2H4) = 0.314 mol N2H4
Step 3: Determine the limiting reactant.
Compare the number of moles of N2H4 (0.314 mol) and H2O2 (1.99 mol).
Since 0.314 mol < 1.99 mol, N2H4 is the limiting reactant.
Step 4: Determine the theoretical yield of HNO3 using the limiting reactant.
Mole ratio: 7 moles H2O2 / 1 mole N2H4
0.314 mol N2H4 * (7 mol H2O2 / 1 mol N2H4) = 2.198 mol H2O2
Step 5: Convert moles of HNO3 to grams.
Mole ratio: 2 moles HNO3 / 7 moles H2O2
2.198 mol H2O2 * (2 mol HNO3 / 7 mol H2O2) = 0.628 mol HNO3
0.628 mol HNO3 * (63.02 g HNO3 / 1 mol HNO3) = 39.57 g HNO3
Finally, since the yield of the reaction is given as 76.5%, we calculate the actual yield as:
0.765 * 39.57 g HNO3 = 27.4 g HNO3
So, the answer to question c is 27.4 grams of HNO3.