This is an oxidation-reduction (redox) reaction:
Na0 - 1 e- → NaI (oxidation)
2 HI + 2 e- → 2 H0 (reduction)
O-II - 1 e- → O-I (oxidation)
Na is a reducing agent, H2O is an oxidizing agent, H2O is a reducing agent.
Input interpretation
Na + H2O → NaO + H2
Sodium water hydrogen
Balanced equation
Balance the chemical equation algebraically:
Na + H2O → NaO + H2
Add stoichiometric coefficients, c;, to the reactants and products:
c1 Na + c2 H2O → c3NaO + c4 H2
Set the number of atoms in the reactants equal to the number of atoms in the products for Na, H and O:
Na: c1 =c3
H: 2c2=2c4
O: c2=c3
Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c1 = 1 and solve the system of equations for the remaining coefficients:
c1 = 1 c2=1
c3=1 c4=1
Substitute the coefficients into the chemical reaction to obtain the balanced equation:
Answer:
Na + H2O → NaO + H2
