A radioactive isotope of vanadium, V, decays by producing a particle and gamma ray. The nuclide formed has the atomic number:
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Answer:
1.57 mol NaN₃
Explanation:
- 2 NaN₃ (s) → 2 Na (s) + 3 N₂ (g)
First we <u>use PV=nRT to calculate the number of N₂ moles that need to be produced</u>:
- P = 1.07 atm
- V = 53.4 L
- n = ?
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 23.7 °C ⇒ 23.7 + 273.16 = 296.86 K
<u>Inputing the data</u>:
- 1.07 atm * 53.4 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.86
And <u>solving for n</u>:
- n = 2.35 mol N₂
Finally we <u>convert N₂ moles into NaN₃ moles</u>, using <em>the stoichiometric coefficients of the balanced reaction</em>:
- 2.35 mol N₂ *
= 1.57 mol NaN₃