A sample of 6.54 g of zinc, with a molar mass of 65.4 g/mol, reacts completely with dilute nitric…
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A sample of of zinc, with a molar mass of , reacts completely with dilute nitric acid, and the gas given off is dinitrogen oxide. The same mass of zinc is then dissolved in hot concentrated acid, and the gas is nitrogen dioxide. Use electron counts: each zinc atom gives electrons, each N2O molecule takes electrons and each NO2 molecule takes electron. Find the volume of each gas at and , taking . Find the moles of nitric acid used in the dilute reaction, from .
Need a reminder? The key ideaShow
Oxidation states of nitrogen run from to . Nitric acid is reduced to NO2 when concentrated and to NO, N2O or NH4+ when dilute. Count electrons: a metal atom that gives up electrons needs electrons to be taken by the nitrogen. At 273 K and 1 atm, 1 mol of gas fills 22.4 L. The Ostwald process gives 1 mol of HNO3 for each mole of NH3 at a full yield.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. The electrons given by zinc equal the electrons taken by the gas.
Step 1. Electrons. of zinc gives of electrons.
Step 2. Dilute acid. Each N2O takes 8 electrons, so of N2O forms. The volume is .
Step 3. Concentrated acid. Each NO2 takes 1 electron, so of NO2 forms, and the volume is .
Step 4. Acid in the dilute reaction. The equation uses acid for zinc, so .
Answer: of N2O, of NO2 and of acid.
Check. The balanced equation gives of NO2, the same answer. The concentrated acid needs , more than the of the dilute reaction.
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