air bags actived when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. this causes NaN3 to decompose explosively according to the following reaction:
2NaN3(s)=2Na(s)+3N2(g)
what mas of NaN3 must be reacted to inflate an air bag to 70.0L at 0C & 1.00 atm?
is this correct:
n(N2): PV/RT
n=1.00 atm x 70.0L/0.0821(L*atm/mol*K) x273K = 3.12 mol N2
mass NaN3 reacted:
(3.12 mol N2) x (2mol NaN3/3molN2)x (65.02gNaN3/1molNaN3)= 135g NaN3
please let me know if I'm doing this correct and where can I find more examples to practice more, have my test tomorrow...thankx. :)
Yes, your calculations are correct for determining the mass of NaN3 that must be reacted to inflate an airbag to 70.0 L at 0°C and 1.00 atm.
To find the number of moles of N2 gas, you correctly used the ideal gas law equation: PV = nRT.
Then, you converted the moles of N2 to moles of NaN3 using the balanced chemical equation, where for every 3 moles of N2, there are 2 moles of NaN3.
Finally, you converted the moles of NaN3 to grams of NaN3 using the molar mass of NaN3, which is 65.02 g/mol.
Your final answer of 135 g NaN3 appears to be correct.
To find more examples to practice similar calculations, you can try looking for additional practice problems in your textbook, lecture notes, or online resources. You can also search for specific topics related to stoichiometry and gas laws to find more practice problems. Good luck with your test!
Yes, your calculations are correct! To find the mass of NaN3 that must be reacted, you followed the correct steps:
1. Calculate the number of moles of N2 using the Ideal Gas Law: n(N2) = PV/RT, where P is the pressure (1.00 atm), V is the volume (70.0 L), R is the ideal gas constant (0.0821 L*atm/mol*K), and T is the temperature (0°C = 273 K).
n(N2) = (1.00 atm) x (70.0 L) / (0.0821 L*atm/mol*K) x (273 K) = 3.12 mol N2
2. Use the balanced equation to calculate the number of moles of NaN3 needed to produce the given number of moles of N2. From the balanced equation, you know that 3 moles of N2 are produced from 2 moles of NaN3.
n(NaN3) = (3.12 mol N2) x (2 mol NaN3 / 3 mol N2) = 2.08 mol NaN3
3. Finally, use the molar mass of NaN3 to convert moles to grams.
mass(NaN3) = (2.08 mol NaN3) x (65.02 g NaN3 / 1 mol NaN3) = 135 g NaN3
Therefore, you correctly found that 135 grams of NaN3 must be reacted to inflate the airbag.
To find more examples to practice similar calculations, you can:
1. Consult your textbook or class notes: Review the chapters or sections that cover stoichiometry and gas laws.
2. Online resources: Look for websites or educational platforms that provide practice problems on stoichiometry and gas laws. Websites like Khan Academy, Chem LibreTexts, and ChemTeam are good sources for practice problems.
3. Practice books or study guides: Check if you have any practice books or study guides that include examples and problems on stoichiometry and gas laws.
Remember to read and understand the given questions carefully and refer to your class materials to ensure accuracy in your calculations. Good luck with your test!