$15/pthh:$
$4Al + 3Cl2 → 2Al2Cl3$
$nAl=2,7/27=0,1mol$
theo pt :
$nCl2=3/4.nAl=3/4.0,1=0,075mol$
$⇒mCl2=5,325g$
$16/$
$nP=3,1/31=0,2mol$
$pthh :$
$4P+5O2→2P2O5$
$a/$
theo pt :
$nO2=5/4.nP=5/4.0,2=0,25mol$
$⇒V_{O_{2}}+0,25/22,4=5,6l$
$b/$
theo pt:
$nP2O5=1/2.nP=1/2.0,2=0,1mol$
$⇒mP2O5=0,1.142=14,2g$