2NaN3(s) -> 2Na(s) + 3N2(g)
What mass of sodium azide should be loaded in the device to ensure full inflation of the same air bag if the actual reaction can be expected to proceed to give only 83% yield of nitrogen gas?
What's with the "same air bag"?
Sorry, I have forgotten to give the information. This question has part a, which is "What mass of sodium azide must be completely reacted to inflate an air bag to 65.0 liters at 20.0 ºC and 1.06 atm?"
Please help me because I have calculated many time, but still cannot get the correct answer, thank you.
To determine the mass of sodium azide needed for full inflation of the airbag, we need to consider the stoichiometry of the reaction and the yield of nitrogen gas.
From the balanced equation:
2NaN3(s) -> 2Na(s) + 3N2(g)
We can see that for every 2 moles of NaN3, we obtain 3 moles of N2 gas.
To calculate the amount of NaN3 needed, we can use the following steps:
1. Determine the number of moles of N2 gas needed.
Since we want the airbag to be fully inflated, we need 3 moles of N2 gas.
2. Adjust for the yield of nitrogen gas.
Since the reaction only has a yield of 83% for N2 gas, we need to take this into account.
Actual moles of N2 produced = Yield (%) * Moles of N2 needed
Actual moles of N2 produced = 0.83 * 3 moles
Actual moles of N2 produced = 2.49 moles
3. Convert moles of N2 to moles of NaN3.
From the stoichiometry of the reaction, we know that for every 3 moles of N2, we need 2 moles of NaN3.
Moles of NaN3 needed = Moles of N2 * (2 moles NaN3 / 3 moles N2)
Moles of NaN3 needed = 2.49 moles * (2 moles NaN3 / 3 moles N2)
Moles of NaN3 needed = 1.66 moles
4. Calculate the mass of NaN3 needed.
To calculate the mass, we need to use the molar mass of NaN3, which is 65 grams/mol (sodium's atomic mass is 23 grams/mol, and nitrogen's atomic mass is 14 grams/mol).
Mass of NaN3 needed = Moles of NaN3 * Molar mass of NaN3
Mass of NaN3 needed = 1.66 moles * 65 grams/mol
Mass of NaN3 needed = 107.9 grams
Therefore, approximately 107.9 grams of sodium azide should be loaded in the device to ensure full inflation of the airbag, taking into account the 83% yield of nitrogen gas.
To answer this question, we need to determine the amount of sodium azide required based on the amount of nitrogen gas produced. We know that the reaction has a yield of 83%, meaning that only 83% of the expected nitrogen gas is produced.
Let's assume the desired amount of nitrogen gas is \(M\) grams. Since there is a 1:3 molar ratio between sodium azide (NaN3) and nitrogen gas (N2), we can calculate the molar mass of sodium azide (65 grams/mol) and use stoichiometry to find the amount of sodium azide required.
Step 1: Calculate the moles of nitrogen gas required
Using the molar mass of nitrogen gas (28 grams/mol), we can find the moles of nitrogen gas required by dividing the desired mass by the molar mass:
\(N_{2}\) moles = \(M\, grams / 28\, grams/mol\)
Step 2: Calculate the moles of sodium azide required
Since the molar ratio of sodium azide to nitrogen gas is 2:3, we can set up a proportion to find the moles of sodium azide required:
\(Moles\, NaN_{3} / 2 = Moles\, N_{2} / 3\)
\(Moles\, NaN_{3} = (2/3) * Moles\, N_{2}\)
Step 3: Adjust for the expected yield
Since the actual reaction has a yield of 83%, we need to adjust the moles of sodium azide required to reflect this:
\(Adjusted\, Moles\, NaN_{3} = Moles\, NaN_{3} / (0.83)\)
Step 4: Calculate the mass of sodium azide required
Finally, we can find the mass of sodium azide required by multiplying the adjusted moles by the molar mass of sodium azide:
Mass of sodium azide required = Adjusted Moles \(NaN_{3}\) * Molar mass of \(NaN_{3}\) (65 grams/mol)
By following these steps, you should be able to calculate the mass of sodium azide required to ensure full inflation of the airbag when considering the yield of the reaction.