If 10.3 g of mg3n2 is treated with water, what volume of gas would be collected at stp

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.

What does that mean you dum bum