The energy used to power one of the Apollo lunar missions was supplied by the following overall reaction:
2N2H4 + (CH3)2N2H2 + 3N2O4 -> 6N2 + 2CO2 + 8H2O
For the phase of the lunar module ascended from the surface of the moon, a total of 1200kg N2H4 was available to react with 1000kg (CH3)2N2H2 and 4500kg N2O4.
a. For this portion of the flight, which of the allocated components was used up first?
b. How much water, in kg, was put into the lunar atmosphere through this reaction?
Answers:
a. N2O4
b. 2.350x10^3kg H2O
I don't know how to get these answers everytime I do it I get a really big number
I work these by working three "simple" stoichiometry problems and taking the smallest value of moles H2O produced.
For example, for N2H4, it is
kmoles N2H4 = 1200/30.029= 39.96
kmoles H2O = 39.96*(8/2) = 159.85
Do the same for N2O4 and (CH3)2N2H2 and take the smaller kmoles, then
kg H2O = kmoles x 18.015 = ??
The answers you posted are correct.
Remember to round to the correct number of significant figures.
Well, here's something to lighten the mood: Why did the astronaut bring a ladder to the moon? Because he wanted to take a step up in life! 😄
Now, let's tackle your questions:
a. Which allocated component was used up first?
To determine this, we need to compare the stoichiometric coefficients of the reactants. Looking at the reaction equation, we see that it takes 3 moles of N2O4 to react with 2 moles of N2H4. Therefore, the ratio of N2O4 to N2H4 is 3:2.
Given that 1200kg of N2H4 was available, we can calculate the amount of N2O4 we would need by using the ratio:
(1200kg N2H4) * (3/2) = 1800kg N2O4
Since only 4500kg of N2O4 was available, which is less than 1800kg, we can conclude that N2O4 was the component used up first.
b. How much water was produced?
Looking at the balanced equation, we see that for every 8 moles of N2H4 reacted, 8 moles of H2O are produced. Therefore, the ratio of water to N2H4 is 1:1.
Given that 1200kg of N2H4 was used up, the amount of water produced would be the same:
1200kg H2O
So, the amount of water put into the lunar atmosphere through this reaction would be 1200kg.
I hope that helps! If you have more questions, feel free to ask.
To determine which component was used up first and calculate the amount of water produced, we need to perform some calculations. Let's break it down step by step:
Given:
- N2H4 available: 1200 kg
- (CH3)2N2H2 available: 1000 kg
- N2O4 available: 4500 kg
Step 1: Determine the limiting reactant
To find out which component is used up first, we need to compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation.
The balanced equation is:
2N2H4 + (CH3)2N2H2 + 3N2O4 -> 6N2 + 2CO2 + 8H2O
First, we need to convert the given masses into moles. To do this, we divide the mass (in kg) of each component by its molar mass.
Molar masses:
- N2H4: 32 g/mol
- (CH3)2N2H2: 74 g/mol
- N2O4: 92 g/mol
Number of moles:
- N2H4: (1200 kg) / (32 g/mol) = 37500 mol
- (CH3)2N2H2: (1000 kg) / (74 g/mol) = 13514 mol
- N2O4: (4500 kg) / (92 g/mol) = 48913 mol
Now, we compare the ratio of the number of moles of each component to their stoichiometric coefficients:
- N2H4 / 2 = 37500 mol / 2 = 18750 mol
- (CH3)2N2H2 = 13514 mol
- N2O4 / 3 = 48913 mol / 3 = 16304 mol
The lowest ratio corresponds to (CH3)2N2H2, which means it is the limiting reactant.
Answer to question a: The (CH3)2N2H2 component is used up first.
Step 2: Calculate the amount of water produced
From the balanced equation, we know that for every 8 moles of water (H2O) produced, we need 1 mole of (CH3)2N2H2.
Since (CH3)2N2H2 is the limiting reactant, it will be completely consumed, producing the maximum amount of water possible.
Given that only 1 mole of (CH3)2N2H2 produces 8 moles of water, we can calculate the amount of water produced as follows:
Amount of water produced = (13514 mol) * (8 mol H2O / 1 mol (CH3)2N2H2) = 108,112 mol H2O
To convert this to kilograms, we multiply by the molar mass of water (18 g/mol) and then convert grams to kilograms:
Amount of water produced = (108112 mol) * (18 g/mol) * (1 kg / 1000 g) = 1946 kg
Answer to question b: The reaction produces approximately 1946 kg of water.
Note: The given answer of 2.350x10^3 kg is incorrect; the accurate value is 1946 kg.
To determine which component was used up first in the reaction, we need to calculate the number of moles of each reactant available and compare them to the stoichiometry of the reaction. The reactant that has the least number of moles compared to the stoichiometry will be used up first.
Let's start by calculating the number of moles of each reactant:
1. N2H4:
Mass of N2H4 = 1200 kg
Molar mass of N2H4 = 32.045 g/mol
Number of moles of N2H4 = (1200 kg) / (32.045 g/mol) = 37447.6 mol
2. (CH3)2N2H2:
Mass of (CH3)2N2H2 = 1000 kg
Molar mass of (CH3)2N2H2 = 58.1 g/mol
Number of moles of (CH3)2N2H2 = (1000 kg) / (58.1 g/mol) = 17213.9 mol
3. N2O4:
Mass of N2O4 = 4500 kg
Molar mass of N2O4 = 92.02 g/mol
Number of moles of N2O4 = (4500 kg) / (92.02 g/mol) = 48,935.4 mol
Next, let's examine the stoichiometry of the reaction:
2N2H4 + (CH3)2N2H2 + 3N2O4 -> 6N2 + 2CO2 + 8H2O
From the balanced equation, we can see that the ratio of N2H4 to N2O4 is 2:3. Therefore, for every 3 moles of N2O4, we need 2 moles of N2H4.
Now, let's calculate the moles of N2O4 that would react with the available N2H4:
Moles of N2O4 needed = (2/3) * Moles of N2H4 = (2/3) * 37447.6 mol = 24965.1 mol
Comparing this to the actual moles of N2O4 present (48,935.4 mol), we can see that the allocated amount of N2O4 is greater than what is needed. Therefore, the N2O4 will be used up first.
To calculate the amount of water (H2O) produced, we need to determine the moles of N2H4 that reacted based on the stoichiometry.
From the balanced equation, we can see that for every 8 moles of H2O produced, we need 2 moles of N2H4.
Moles of H2O produced = (8/2) * Moles of N2H4 = (8/2) * 37447.6 mol = 149,790.4 mol
Next, we can calculate the mass of water produced:
Molar mass of H2O = 18.015 g/mol
Mass of H2O produced = (149,790.4 mol) * (18.015 g/mol) = 2,694,451.6 g = 2,694.4516 kg
Therefore, the amount of water produced through this reaction is approximately 2,694.45 kg.
Please note that the numbers given may have been rounded for simplification, which could account for slight variations in the final answer.