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?
Find the molar mass of N2H2. Now find the number of moles that are in 5.0 x 10^14g of N2H2. With me so far?
Oh, sorry, I thought I got to your question earlier than I did. I guess this site doesn't work too well for real time discussions...
how do i find molar mass?
molar mass = the sum of the atomic masses which you get from the periodic table.
To solve this problem, we need to use the balanced chemical equation and the given information.
1) To determine the number of moles of nitrogen (N2) that can be created, we need to use the stoichiometric coefficients from the balanced equation. According to the equation, 2 moles of hydrazine (N2H4) react to produce 3 moles of nitrogen (N2). Therefore, the ratio of moles of nitrogen to moles of hydrazine is 3/2.
Given that you have 5.0 x 10^4 grams of hydrazine (N2H4):
First, we need to convert the mass of hydrazine to moles using its molar mass.
The molar mass of N2H4 = (2 x 14.01 g/mol) + (4 x 1.01 g/mol) = 32.05 g/mol
Moles of N2H4 = Mass of N2H4 / Molar mass of N2H4
Moles of N2H4 = 5.0 x 10^4 g / 32.05 g/mol ≈ 1,559.82 mol (rounded to four decimal places)
Now, we can use the stoichiometric ratio to determine the moles of nitrogen:
Moles of N2 = Moles of N2H4 x (3/2)
Moles of N2 = 1,559.82 mol x (3/2) ≈ 2,339.73 mol (rounded to four decimal places)
Therefore, approximately 2,339.73 moles of nitrogen (N2) can be created.
2) Using the same approach, we can determine the moles of water (H2O) produced. According to the balanced equation, for every 2 moles of hydrazine used, 4 moles of water are produced. Thus, the ratio of moles of water to moles of hydrazine is 4/2 = 2/1.
Moles of H2O = Moles of N2H4 x (2/1)
Moles of H2O = 1,559.82 mol x (2/1) ≈ 3,119.64 mol (rounded to four decimal places)
Therefore, approximately 3,119.64 moles of water (H2O) can be produced.
3) To determine the mass and volume of dinitrogen tetroxide (N2O4) required, we consider the stoichiometric ratio between N2O4 and N2H4. According to the balanced equation, 1 mole of N2O4 is required for every 2 moles of N2H4. Therefore, the ratio of moles of N2O4 to moles of N2H4 is 1/2.
Given that density = 1.44 g/cm^3, we can calculate the required mass of N2O4 using its molar mass.
Molar mass of N2O4 = (2 x 14.01 g/mol) + (4 x 16.00 g/mol) = 92.01 g/mol
Mass of N2O4 = Moles of N2H4 x (1/2) x Molar mass of N2O4
Mass of N2O4 = 1,559.82 mol x (1/2) x 92.01 g/mol ≈ 71,610.36 g (rounded to two decimal places)
Therefore, approximately 71,610.36 grams of N2O4 is needed.
To calculate the volume of N2O4, we can use its density.
Density of N2O4 = Mass of N2O4 / Volume of N2O4
Rearranging the equation: Volume of N2O4 = Mass of N2O4 / Density of N2O4
Volume of N2O4 = 71,610.36 g / 1.44 g/cm^3 ≈ 49,778.75 cm^3 (rounded to two decimal places)
Therefore, approximately 49,778.75 cm^3 (or 49.78 L) of N2O4 is needed.
Note: It is always essential to round the calculated values to the appropriate number of significant figures based on the given data.