1 N204(l) + 2 N2H4(l) --> 3 N2(g) + 4 H2O(g)
If you start with: 5.0 x 10^4 g hydrazine (N2H4) "in the tank":
1) How many moles of nitrogen can be created (assuming 100% yield)?
2) How many moles of water can be produced?
2) What mass of dinitrogen tetroxide (N2O4) is needed? What volume of N2O4 is needed? (density = 1.44 g/cm^3)
Each of your 1,2,3 questions are separate stoichiometry problems. Here is an example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Post your work if you get stuck.
To answer these questions, we need to use stoichiometry, which relates the amounts of reactants and products in a balanced chemical equation. The given balanced equation is:
1 N204(l) + 2 N2H4(l) --> 3 N2(g) + 4 H2O(g)
1) To determine the number of moles of nitrogen (N2) that can be produced, we need to use the mole ratio between N2 and N2H4. From the balanced equation, we can see that 2 moles of N2H4 produce 3 moles of N2. So, the mole ratio is 3 moles of N2 per 2 moles of N2H4.
To calculate the number of moles of nitrogen from the given mass of hydrazine (N2H4), we need to know the molar mass of N2H4. The molar mass of N2H4 is (2 * atomic mass of N) + (4 * atomic mass of H) = (2 * 14.01 g/mol) + (4 * 1.01 g/mol) = 32.05 g/mol.
Given that there are 5.0 x 10^4 g of hydrazine (N2H4), we can now calculate the number of moles:
Number of moles = mass / molar mass
Number of moles = 5.0 x 10^4 g / 32.05 g/mol
Number of moles = 1560.43 mol (rounded to three significant figures)
Since the stoichiometry ratio is 3 moles of N2 per 2 moles of N2H4, we can calculate the number of moles of N2 produced:
Number of moles of N2 = (1560.43 mol N2H4 * 3 mol N2) / 2 mol N2H4
Number of moles of N2 = 2340.65 mol (rounded to three significant figures)
Therefore, if the hydrazine (N2H4) is completely converted to nitrogen (N2), we can produce 2340.65 moles of nitrogen.
2) To determine the number of moles of water (H2O) produced, we need to use the mole ratio between H2O and N2H4. From the balanced equation, we can see that 2 moles of N2H4 produce 4 moles of H2O. So, the mole ratio is 4 moles of H2O per 2 moles of N2H4.
Using the same number of moles of N2H4 (1560.43 mol) as calculated in part 1, we can calculate the number of moles of H2O produced:
Number of moles of H2O = (1560.43 mol N2H4 * 4 mol H2O) / 2 mol N2H4
Number of moles of H2O = 3120.86 mol (rounded to three significant figures)
Therefore, if the hydrazine (N2H4) is completely converted to water (H2O), we can produce 3120.86 moles of water.
3) To determine the mass of dinitrogen tetroxide (N2O4) needed, we need to use the mole ratio between N2O4 and N2H4. From the balanced equation, we can see that 1 mole of N2O4 is produced for every 2 moles of N2H4. So, the mole ratio is 1 mole of N2O4 per 2 moles of N2H4.
Given that there are 1560.43 moles of N2H4, we can calculate the number of moles of N2O4:
Number of moles of N2O4 = 1560.43 mol N2H4 / 2 mol N2H4
Number of moles of N2O4 = 780.22 mol (rounded to three significant figures)
Now, to determine the mass of N2O4 needed, we use the molar mass of N2O4, which is (2 * atomic mass of N) + (4 * atomic mass of O) = (2 * 14.01 g/mol) + (4 * 16.00 g/mol) = 92.02 g/mol.
Mass of N2O4 needed = number of moles of N2O4 * molar mass of N2O4
Mass of N2O4 needed = 780.22 mol * 92.02 g/mol
Mass of N2O4 needed = 71,768.79 g (rounded to three significant figures)
Therefore, we would need approximately 71,768.79 grams of dinitrogen tetroxide (N2O4) for this reaction.
To determine the volume of N2O4 needed, we can use its density. Given that the density of N2O4 is 1.44 g/cm^3, we will convert the mass of N2O4 to volume.
Volume of N2O4 needed = mass of N2O4 needed / density of N2O4
Volume of N2O4 needed = 71,768.79 g / 1.44 g/cm^3
Volume of N2O4 needed = 49,814.72 cm^3 (rounded to three significant figures)
Therefore, we would need approximately 49,814.72 cm^3 or milliliters (mL) of dinitrogen tetroxide (N2O4) for this reaction.