3. a.) The initial reaction when an automobile airbag deploys is: 2NaN3(s) → 2Na(s) + 3N2(g). If 125.0g of NaN3 reacts, how many liters of N2 gas result at 25.0oC and a pressure of 1.00 atmosphere?
3. b.) The recent death of a former soviet spy is attributed to the effects of an isotope of Polonium with an atomic number of 84 and a mass number of 210. This isotope is an alpha emitter. What are the atomic number, mass number and name of the element produced when one of the polonium atoms emits an alpha particle?
3. c.) Polonium 210 has a half life of 138 days. If an initial sample of it has a mass of 1.50 grams, how many grams of it remain after:
A. 138 days ? B. 2(138) = 376 days ? C. 3(138) = 414 days ?
The initial reaction when an automobile airbag deploys is:
2 NaN3(s) → 2 Na(s) + 3 N2(g).
If 125.0 g of NaN3 reacts, how many liters of N2 gas result at 25.0 degrees C and a pressure of 1.00 atmosphere?
Let us first convert 125.0 g of NaN3 into moles of NaN3. We do this by first working out the molar mass of NaN3. It is 65.01 g/mol.
(1 mol/65.01 g)(125.0 g) = 1.923 mol
Therefore, we react 1.923 mol of NaN3. Now, we set up a little ratio to figure out moles of N2 produced. From the stoichiometry of the balanced chemical equation, we know that 2 moles of NaN3 react to produce 3 moles of N2.
Therefore, 2 is to 3 as 1.923 is to x.
2/3 = 1.923/x
x = 2.88
This solution provides detailed answers to three questions regarding atmosphere, atomic weight, and mass.