(a)(i) Define heat of neutralization (ii) Give the reason why copper (II) chloride can be prepared by neutralization, unlike lead (II) chloride. (b)(i) Describe in outline, the manufacture of trioxonitrate (V) acid by the catalytic oxidation of ammonia, giving equations where appropriate. (ii) What are the products obtained when sodium tioxonitrate (V) is heated strongly?
(c) When powdered magnesium is heated to redness in a stream of nitrogen, magnesium nitride (Mg\(_3\)N\(_2\)) is formed. (i) Write an equation for the reaction (ii) Hence, calculate the amount (in mole) of magnesium nitride that can be obtained from 3.0g of magnesium [Mg = 24].
Explanation
(a)(i) Heat of neutralization is defined as the amount of heat evolved when one mole of hydrogen ions (H\(^+\)/H\(^3\)O\(+\)) from an acid reacts with one mole of hydroxide ions (OH\(^-\)) from an alkali to form one mole of water. (ii) Neutralization can be used to prepare only soluble salts, hence CuCl\(_2\) which is soluble can be prepared by this method while PbC\(_2\) which is insoluble cannot.
(b)(i) Manufacture of HNO\(_3\) in outline by the catalytic oxidation of NH\(_3\). Ammonia reacts with air or oxygen in the presence of platinum-rhodium catalyst and a high temperature (of about 550 - 850°C) to form Nitrogen (11) oxide and steam. 4NH\(_3\) + 5O\(_3\) ---> 4NO + 6H\(_2\)O. Nitrogen (II) oxide reacts with more or excess air to give nitrogen (IV) oxide which is dissolved' in hot water in the presence of more air to give trioxonitrate (V) acid. 2NO + O\(_2\) —> 2NO\(_2\); 4NO\(_2\) + 2H\(_2\)O + O\(_2\) —> 4HNO\(_3\) (ii) Oxygen, NaNO\(_3\) are obtained when NaNO\(_3\) is heated strongly.
(c) Reactions of Mg with nitrogen. (i) Equations of the reaction 3Mg + N\(_2\) --> Mg\(_3\)N\(_2\). (ii) Calculation of the amount of Mg\(_3\)N\(_2\) that can be obtained from 3.0g of Mg. From the equation, (3 x 24)g of Mg gives 1 mole of Mg\(_3\)N\(_2\). 3g of Mg will give 3 x \(\frac{1}{3 \times 24}\) mole of Mg\(_3\)N\(_2\) = 0.0417 mole.
