What will be the volume, at 115 kPa and 355K, of the nitrogen from decomposition of 35.8 g of sodium azide, NaN3? 2NaN3 -> 2Na + 3N2
2007-05-31 03:27:01 · 2 answers · asked by violenta10906 1 in Science & Mathematics ➔ Chemistry
Just combine what you know about stoichiometry with the ideal gas law.
Convert the mass of sodium azide into moles. Then, using stoichiometry, calculate the moles of N2 that can be produced.
Next, use the ideal gas law to calculate the volume of that many moles of N2 gas.
2007-05-31 03:32:07 · answer #1 · answered by hcbiochem 7 · 0⤊ 1⤋
Mr of sodium azide =65g/mol
moles of NaN3 = 35.8g / 65 =0.5508 mol
moles of N2 = 0.5508 mol x 3/2 = 0.8262 mol
R =8.314J/mol.K
T=355K
P = 115000Pa
Sub all the values into the gas equation, PV=nRT
V = 0.0212m3 =21.2L
2007-05-31 10:36:27 · answer #2 · answered by 1-man-show 3 · 1⤊ 0⤋