Question

Складіть рівняння реакцій горіння складних речовин, якщо в усіх випадках утворюються оксиди двох елементів з максимально можливою валентністю. в)...C2H4+...O2=г)...C2H6O+..O2=д)...PH+...O2=​

Answer 1

Ответ:в) C2H4 + 3O2 → 2CO2 + 2H2O

г) C2H6O + 3O2 → 2CO2 + 3H2O

д) 4PH3 + 8O2 → 2P2O5 + 6H2O

Answer 2

Equations for the Combustion of Complex Substances

When complex substances undergo combustion, they react with oxygen to form oxides of the elements involved. The equations for the combustion of the given substances are as follows:

a) C2H4 + 3O2 → 2CO2 + 2H2O [[1]] b) C2H6O + 3O2 → 2CO2 + 3H2O [[2]] c) PH + 5O2 → P2O5 + H2O [[3]]

Let's break down each equation:

a) The combustion of ethene (C2H4) produces carbon dioxide (CO2) and water (H2O). The balanced equation shows that 1 mole of C2H4 reacts with 3 moles of O2 to produce 2 moles of CO2 and 2 moles of H2O.

b) The combustion of ethanol (C2H6O) also produces carbon dioxide (CO2) and water (H2O). In this case, 1 mole of C2H6O reacts with 3 moles of O2 to produce 2 moles of CO2 and 3 moles of H2O.

c) The combustion of phosphine (PH) produces diphosphorus pentoxide (P2O5) and water (H2O). Here, 1 mole of PH reacts with 5 moles of O2 to produce 1 mole of P2O5 and 1 mole of H2O.

These equations represent the combustion reactions of the given substances, where the resulting oxides have the maximum possible valence states for the elements involved.

I hope this helps! Let me know if you have any further questions.