Расставьте коэффициенты в следующих схемах реакций:
a) $Fe + O_{2} ⟶ Fe_{3}O_{4}$;
$S + O_{2} ⟶ SO_{2}$;
$CH_{4} ⟶ C + H_{2}$;
$P + Br_{2} ⟶ PBr_{5}$;
б) $Al + F_{2} ⟶ AlF_{3}$;
$N_{2} + H_{2} ⟶ NH_{3}$;
$FeO + O_{2} ⟶ Fe_{3}O_{4}$;
$Fe + Cl_{2} ⟶ FeCl_{3}$;
в) $H_{2}O_{2} ⟶ H_{2}O + O_{2}$;
$Ca + O_{2} ⟶ CaO$;
$Na + Cl_{2} ⟶ NaCl$;
$BaO + O_{2} ⟶ BaO_{2}$;
г) $CuO ⟶ Cu_{2}O + O_{2}$;
$SO_{2} + O_{2} ⟶ SO_{3}$;
$FeBr_{3} ⟶ FeBr_{2} + Br_{2}$;
$CuCl_{2} + Na_{2}S ⟶ CuS + NaCl$;
д) $H_{2}S + O_{2} ⟶ SO_{2} + H_{2}O$;
$C_{4}H_{10} + O_{2} ⟶ CO_{2} + H_{2}O$;
$Ag + H_{2}S + O_{2} ⟶ Ag_{2}S + H_{2}O$;
$CH_{5}N + O_{2} ⟶ CO_{2} + H_{2}O + N_{2}$;
е) $Zn + HCl ⟶ ZnCl_{2} + H_{2}$;
$KOH + H_{2}SO_{4} ⟶ K_{2}SO_{4} + H_{2}O$;
$NaOH + H_{3}PO_{4} ⟶ Na_{3}PO_{4} + H_{2}O$;
$CuOH ⟶ Cu_{2}O + H_{2}O$.
$3Fe + 2O_{2} = Fe_{3}O_{4}$
$S + O_{2} = SO_{2}$
$CH_{4} = C + 2H_{2}$
$2P + 5Br_{2} = 2PBr_{5}$
$2Al + 3F_{2} = 2AlF_{3}$
$N_{2} + 3H_{2} = 2NH_{3}$
$6FeO + O_{2} = 2Fe_{3}O_{4}$
$2Fe + 3Cl_{2} = 2FeCl_{3}$
$2H_{2}O_{2} = 2H_{2}O + O_{2}$
$2Ca + O_{2} = 2CaO$
$2Na + Cl_{2} = 2NaCl$
$2BaO + O_{2} = 2BaO_{2}$
$4CuO = 2Cu_{2}O + O_{2}$
$2SO_{2} + O_{2} = 2SO_{3}$
$2FeBr_{3} = 2FeBr_{2} + Br_{2}$
$CuCl_{2} + Na_{2}S = CuS + 2NaCl$
$2H_{2}S + 3O_{2} = 2SO_{2} + 2H_{2}O$
$2C_{4}H_{10} + 13O_{2} = 8CO_{2} + 10H_{2}O$
$4Ag + 2H_{2}S + O_{2} = 2Ag_{2}S + 2H_{2}O$
$4CH_{5}N + 9O_{2} = 4CO_{2} + 10H_{2}O + 2N_{2}$
$Zn + 2HCl = ZnCl_{2} + H_{2}$
$2KOH + H_{2}SO_{4} = K_{2}SO_{4} + 2H_{2}O$
$3NaOH + H_{3}PO_{4} = Na_{3}PO_{4} + 3H_{2}O$
$2CuOH = Cu_{2}O + H_{2}O$
