When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.
NO2→NO^−3
NO2→NO
and do the usual changes
First, balance the two half reactions:
3. NO2 +H2O →NO^−3 + 2 H^+ + e−
4. NO2 +2 H^+ + 2e− → NO + H2O
Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:
5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−
Now add Eqn 4 and 5 (the electrons now cancel each other):
3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+
and cancel terms that’s common to both sides:
3NO2 + H2O → NO + 2NO^−3 + 2H+
This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.
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Answer:
2.2 moles of Fe will be produced
Explanation:
Step 1: Data given
Number of moles of hydrogen gas = 3.3 moles
Number of moles of iron oxide = 1.5 moles
Step 2: The balanced equation
3H2 + Fe2O3 → 2Fe + 3H2O
Step 3: Calculate the limiting reactant
For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O
Hydrogen gas is the limiting reactant. It will completely be consumed (3.3 moles). Fe2O3 is in excess. There will react 3.3 / 3 = 1.1 moles
There will remain 1.5 - 1.1 = 0.4 moles Fe2O3
Step 4: Calculate moles Fe
For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O
For 3.3 moles H2 we'll have 2/3 * 3.3 = 2.2 moles Fe
2.2 moles of Fe will be produced
