Mg 3 моль + 2HCl = MgCl2 V (H 2) 22,4 67,2 л n( H2 )= 3 моль 22,4 л / моль m( Mg ) n( Mg ) * M ( Mg ) m(Mg)=3*24=72 г Ответ: 72 г n(H2)= + H2 3 моль
You can add this document to your study collection(s)
You can add this document to your saved list
(For complaints, use another form )
Input it if you want to receive answer
Rate us