C3H8(g)+5O2(g)-->3CO2(g)+4H2O(l) what reactant is oxidized and which is reduced?
To determine which reactant is oxidized and which is reduced in the given chemical equation, we need to look at the changes in the oxidation states of the elements involved.
In the reaction:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
Let's consider the oxidation states of carbon (C) and hydrogen (H) in the reactants and products:
- In C3H8, the oxidation state of carbon is +3, while hydrogen is -1.
- In CO2, the oxidation state of carbon is +4, and there are no hydrogen atoms.
- In H2O, the oxidation state of hydrogen is +1, and there are no carbon atoms.
From this analysis, we can see that the oxidation state of carbon in C3H8 has increased from +3 to +4. This means that carbon has been oxidized.
The oxidation state of hydrogen in C3H8 has changed from -1 to +1 in H2O. This indicates that hydrogen has been reduced.
Thus, in the given reaction, propane (C3H8) is oxidized, and oxygen (O2) is reduced.