Daca in reactie se folosesc 124 g P4 calculati volumul de oxigen in conditii normale care se consuma si masa de pentaoxid de fosfor ce rezulta din reactie.
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1mol..5mol...2mol
P4 + 5O2 = 2P2O5
1mol...5mol....2mol
n=V/Vm=>V=5*22,4=112L O2
MP2O5=142g/mol ⇒ m=2*142=284g P2O5