An aqueous solution of ammonium nitrite, NH4NO2, decomposes when heated to give off nitrogen, N2.
NH4NO2(s) ==> 2H2O(g) + N2(g)
This reaction may be used to prepare pure nitrogen. How many grams of ammonium nitrite must have reacted if 3.61 dm3 of nitrogen gas was collected over water at 24C and 96.5 kPa?
2006-10-13 04:51:58 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
ideal gas law: PV = nRT
P = 96.5 kPa
V = 3.61 dm^3
n = number of moles
R = Gas constant
T = 24°C
n = PV / (RT)
n = 96.5 kPa * 3.61 dm^3 / ((8.314 J/mol*°K) * 297 °K) = 0.1411 mol
Molecular weight of NH4NO2 = 14+1*4+14+16*2 = 64 g/mol
m = 0.1411 mol * 64 g/mol = 9.03 g NH4NO2
2006-10-13 04:59:16 · answer #1 · answered by MadScientist 4 · 0⤊ 0⤋
As I said..................
2006-10-13 13:57:47 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋