Balanced Chemical Equation:
1 N204(l) + 2 N2H4(l) --> 3 N2(g) + 4 H20(g)
If you start with: 5.0 x 10^4g hydrazine (N2H4) "in the tank":
1) How many moles of nitrogen can be created (assuming 100% yielded)?
2) How many moles of water can be produced?
3) What mass of dinitrogen tetroxide (N2O4) is needed? What volume of N2O4 is needed? (density = 1.44 g/cm^3)
Help please?
Here is an example problem on stoichiometry. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To answer these questions, we need to use the balanced chemical equation given:
1 N2O4(l) + 2 N2H4(l) --> 3 N2(g) + 4 H2O(g)
Let's solve each part of the question step by step:
1) How many moles of nitrogen can be created (assuming 100% yield)?
From the balanced equation, we can see that 1 mole of N2O4 reacts with 2 moles of N2H4 to produce 3 moles of N2. Therefore, the ratio of N2 to N2H4 is 3:2.
To calculate the number of moles of nitrogen that can be created, we need to convert the given mass of hydrazine (N2H4) to moles using its molar mass.
The molar mass of N2H4 is:
(2 * 14.01 g/mol) + (4 * 1.01 g/mol) = 32.05 g/mol
Now, we can calculate the number of moles of N2H4:
moles of N2H4 = mass of N2H4 / molar mass of N2H4
= 5.0 x 10^4 g / 32.05 g/mol
= 1560.1 mol (approximately)
Using the 3:2 ratio from the balanced equation, we can calculate the number of moles of nitrogen (N2):
moles of N2 = (3/2) * moles of N2H4
= (3/2) * 1560.1 mol
= 2340.15 mol
Therefore, approximately 2340.15 moles of nitrogen can be created.
2) How many moles of water can be produced?
From the balanced equation, we can see that 2 moles of N2H4 react to produce 4 moles of H2O. Therefore, the ratio of H2O to N2H4 is 2:1.
Using the number of moles of N2H4 calculated earlier (1560.1 mol), we can calculate the number of moles of water (H2O):
moles of H2O = 2 * moles of N2H4
= 2 * 1560.1 mol
= 3120.2 mol
Therefore, approximately 3120.2 moles of water can be produced.
3) What mass of dinitrogen tetroxide (N2O4) is needed? What volume of N2O4 is needed? (density = 1.44 g/cm^3)
To calculate the mass of N2O4, we need to use the density given. The density of N2O4 is 1.44 g/cm^3.
The molar mass of N2O4 is:
(2 * 14.01 g/mol) + (4 * 16.00 g/mol) = 92.02 g/mol
To find the mass of N2O4 needed, we use the formula:
mass = density * volume
Since the density is given in g/cm^3 and we need the volume in cm^3, we can rearrange the formula as:
volume = mass / density
Given that the molar ratio between N2O4 and N2H4 is 1:2, we need half the number of moles of N2H4 for N2O4. Therefore:
moles of N2O4 = moles of N2H4 / 2
≈ 1560.1 mol / 2
≈ 780.05 mol
Now, we can use the formula:
volume = mass / density
= (molar mass of N2O4) * (moles of N2O4) / density
= (92.02 g/mol) * (780.05 mol) / (1.44 g/cm^3)
≈ 49737.39 cm^3 (approximately)
Therefore, approximately 49737.39 cm^3 of N2O4 is needed.
In summary:
1) Approximately 2340.15 moles of nitrogen can be created.
2) Approximately 3120.2 moles of water can be produced.
3) Approximately 49737.39 cm^3 of N2O4 is needed, or equivalently, the mass of N2O4 needed is 49737.39 g.