Respuesta :
Use the ideal gas law: PV=nRt
To find volume, move to V=(nRt)/P
N=# of moles (to find, 10.3 g/100.929g)=.102 moles
R=ideal gas constant(this value is 0.082056(Liters*atmospheres)/(Kelvin*mol)
T= temp which at stp is 273 Kelvin
P= 1 atmosphere
V=((.102 moles)(0.082056 Latm/Kmol)(273 K))/1 atm
Mg3N2 +6H2O → 3Mg(OH)2 +2 NH3
relative weight of Mg3N2 = (3x24)+(2x14) = 100 g/mole
mole of Mg3N2 = 10,3 g/100 g/mole = 0,103 mole
mole of NH3(g) = 2/1 x 0,103 mole = 0,206 mole
volume NH3(g) which collected at STP = mole NH3(g) x 22,4 litre/mole = 0,206 mole x 22,4 litre/mole = 4,6144 litre.
relative weight of Mg3N2 = (3x24)+(2x14) = 100 g/mole
mole of Mg3N2 = 10,3 g/100 g/mole = 0,103 mole
mole of NH3(g) = 2/1 x 0,103 mole = 0,206 mole
volume NH3(g) which collected at STP = mole NH3(g) x 22,4 litre/mole = 0,206 mole x 22,4 litre/mole = 4,6144 litre.