M(Ca3P2)=3*40g+2*31g=182g
M(PH3)=31g+3*1g=34g
Ca3P2+3H2O-->2PH3+3CaO
182g-----3mol
60g-------xmol
x=60g*3mol/182g
x=1mol
1mol<2,5mol, que significa que tenemos el exceso de H2O.
182g Ca3P2 --------------- 34g PH3
60g Ca3P2 ----------------- xg PH3
xg=60g*34g/182g
xg=11,2g PH3