NH4NO2 (s) ----> N2 (g) + H2O (g)
When a sample is decomposed in a test tube, 656 mL of wet N2(g) is collected over water at 28°C and 447 torr total pressure.
How many grams of dry NH4NO2(s) were initially decomposed? The vapor pressure of water at 28°C is 26.3 torr.
Mass? in g
V= 656ml= 0.656L
T= 28°C= 28+273= 301K
447torr= 1/760 x 447= 0.59atm
Vapour pressure of water= 26.3torr= 0.035atm
Partial pressure of N2= 0.59-0.035= 0.555atm
Now let's find n, the number of moles of N2 from ideal gas equation.
n= 0.555 x 0.656/ 0.0831 x 301
From the stoichiometry of reaction
1mol of N2 is produced from 1mol of NH4NO2
0.015mol of N2 will be produced from 0.015mol of NH4NO2
Mole of NH4NO2= 0.015mol
Molar mass of NH4NO2= 64g/mol
Mass of NH4NO2= mole x molar mass= 0.015 x 64= 0.96g of NH4NO2