Here is the question:
Automobile air bags use the decomposition of sodium azide as their source of gas for rapid inflation:
2NaN3(s)-->2Na(s)+3N2(g)
What mass (g) of NaN3 is required to provide 40.0 of N2 at 25.0 C and 763 torr?
Here is my work:
P=1.0039atm or 763torr
V=40.L
R=.0821
T=298.15K
PV/RT=n and my answer comes out to be 65.0099 which I multiply times 3/2 to convert to NaN3 then multiply that number my molecular weight (65.009) to get 6320193.00741
The assessment is saying this answer is incorrect can anyone tell me what I am doing wrong?
There is 2/3 as much NaN3 as N2. Flip your ratio on the conversion.
Once again thank you Bob, I think my brain is fried for the day. I would stop if I did not have 4 more questions to go. Hopefully I can get them on my own now that I figured out the calculator issues I was haveing.
Based on your work, it seems like you are on the right track. However, there might be a calculation error in your final step.
Let's go through the calculation again step by step:
1. Convert the given pressure from torr to atm: 763 torr / 760 torr/atm = 1.0039 atm.
2. Calculate the number of moles of nitrogen gas using the ideal gas law:
PV = nRT
n = PV / RT = (1.0039 atm) * (40.0 L) / (0.0821 (atm L) / (mol K) * 298.15 K) = 1.6090 mol N2.
3. From the balanced equation, we know that 3 moles of N2 are produced from 2 moles of NaN3.
Therefore, we need to double the number of moles of NaN3, which gives us 2 * 1.6090 mol NaN3 = 3.2180 mol NaN3.
4. Finally, calculate the mass of NaN3 using its molar mass. The molar mass of NaN3 is approximately 65.009 g/mol:
Mass = (3.2180 mol NaN3) * (65.009 g/mol) = 209.261 g NaN3.
So, the correct answer is approximately 209.261 grams of NaN3, which is required to produce 40.0 L of N2 at 25.0 °C and 763 torr.
It seems like you are mostly on the right track, but there is a small error in your calculations. Here is the correct step-by-step solution:
1. Convert the given temperature from Celsius to Kelvin:
T = 25.0°C + 273.15 = 298.15 K
2. Convert the given pressure from torr to atm:
P = 763 torr / 760 torr/atm = 1.0039 atm
3. Apply the ideal gas law equation to find the number of moles of N2:
PV = nRT
Plugging in the values:
(1.0039 atm)(40.0 L) = n(0.0821 L·atm/(mol·K))(298.15 K)
Solving for n gives:
n = (1.0039 atm * 40.0 L) / (0.0821 L·atm/(mol·K) * 298.15 K) = 1.612 mol
4. Since the balanced chemical equation shows that 2 moles of NaN3 produce 3 moles of N2, we need to convert the moles of N2 to moles of NaN3.
Using the ratio: 2 mol NaN3 / 3 mol N2
(1.612 mol N2) * (2 mol NaN3 / 3 mol N2) = 1.075 mol NaN3
5. Finally, calculate the mass of NaN3 using the molar mass of NaN3:
Molar mass of NaN3 = 22.990 + 14.007 + (3 * 15.999) = 65.009 g/mol
(1.075 mol NaN3) * (65.009 g/mol) = 69.766 g
Therefore, the correct mass of NaN3 required to provide 40.0 L of N2 at 25.0 °C and 763 torr is approximately 69.766 grams.