7H2O2(aq) + N2H4(g) �¨ 2HNO3(g) + 8H2O(g)
If the rocket engine contains 1500.0 grams of H2O2 and an excess of N2H4, what volume of water vapor will be produced when the rocket fires up to a temperature of 540�‹C and a pressure of 1.2atm?
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To solve this problem, we will use the ideal gas law equation:
PV = nRT
where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
We will need to calculate the number of moles of water vapor produced first:
First, convert the mass of H2O2 to moles using the molar mass of H2O2:
Molar mass of H2O2 = 34.02 g/mol
Number of moles of H2O2 = mass of H2O2 / molar mass of H2O2
= 1500.0 g / 34.02 g/mol
= 44.07 mol
According to the balanced chemical equation, 7 moles of H2O2 produce 8 moles of water vapor:
Number of moles of water vapor = (8/7) * number of moles of H2O2
= (8/7) * 44.07 mol
≈ 50.31 mol
Next, convert the temperature from Celsius to Kelvin:
Temperature in Kelvin = 540 �‹C + 273.15
= 813.15 K
Now, rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
Substitute the known values into the equation:
V = (50.31 mol * 0.0821 L·atm/mol·K * 813.15 K) / 1.2 atm
= 3393.5 L
Therefore, when the rocket fires up, approximately 3393.5 liters of water vapor will be produced.
To find the volume of water vapor produced, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's convert the given temperature from Celsius to Kelvin:
T(K) = 540 + 273.15 = 813.15 K
Next, we need to find the number of moles of water vapor produced using the balanced chemical equation:
7H2O2(aq) + N2H4(g) -> 2HNO3(g) + 8H2O(g)
Since the ratio of H2O2 to H2O is 7:8, we can assume that 1 mole of H2O2 produces 8 moles of H2O. Therefore, we will calculate the moles of water produced based on the number of moles of H2O2.
Given:
Mass of H2O2 = 1500.0 grams
First, calculate the molar mass of H2O2:
H2O2: 2(1.01 g/mol H) + 2(16.00 g/mol O) = 34.02 g/mol
Next, calculate the number of moles of H2O2:
n(H2O2) = mass ÷ molar mass
n(H2O2) = 1500.0 g ÷ 34.02 g/mol ≈ 44.06 mol
Since the balanced chemical equation shows that the stoichiometric coefficient of H2O is 8, the number of moles of H2O produced will also be 8 times the moles of H2O2 used:
n(H2O) = 8 * n(H2O2)
n(H2O) = 8 * 44.06 mol = 352.48 mol
Now, we can use the ideal gas law equation to find the volume of water vapor produced:
PV = nRT
Substituting the values:
P = 1.2 atm
V = volume (to be calculated)
n = 352.48 mol
R = 0.0821 L·atm/(mol·K)
T = 813.15 K
Rearranging the equation to solve for V:
V = nRT ÷ P
Substituting the values:
V = (352.48 mol) * (0.0821 L·atm/(mol·K)) * (813.15 K) ÷ (1.2 atm)
V ≈ 18889.41 L
Therefore, approximately 18889.41 liters of water vapor will be produced when the rocket fires up to a temperature of 540�‹C and a pressure of 1.2 atm.