(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].
(a)(i) Heat of neutralization is the heat change when one mole of hydrogen ions (H+) from an acid reacts completely with one mole of hydroxide ions (OH-) from a base to form one mole of water under standard conditions.
(ii) Copper(II) chloride is soluble in water, so it can be made in solution by neutralizing an acid with a base and then crystallized. Lead(II) chloride is insoluble in cold water, so a neutralization in solution would simply precipitate it; it is therefore prepared by precipitation (double decomposition), not by neutralization.
(b)(i) Manufacture of trioxonitrate(V) acid (Ostwald process)
- Ammonia is mixed with excess air and passed over a platinum-rhodium gauze catalyst at about 900 °C:
\[4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O\]
- The nitrogen(II) oxide is cooled and oxidized by more air:
\[2NO + O_2 \rightarrow 2NO_2\]
- The nitrogen(IV) oxide is dissolved in water in the presence of excess air:
\[4NO_2 + O_2 + 2H_2O \rightarrow 4HNO_3\]
(ii) Sodium trioxonitrate(V) on strong heating gives sodium trioxonitrate(III) and oxygen:
\[2NaNO_3 \rightarrow 2NaNO_2 + O_2\]
(c)(i) \[3Mg + N_2 \rightarrow Mg_3N_2\]
(ii) Amount of Mg \(= \dfrac{3.0}{24} = 0.125\ \text{mol}\).
From the equation, 3 mol Mg give 1 mol Mg3N2, so
amount of Mg3N2 \(= \dfrac{0.125}{3} = 0.0417\ \text{mol} \approx 4.17 \times 10^{-2}\ \text{mol}\).
(a)(i) Heat of neutralization is the heat change when one mole of hydrogen ions (H+) from an acid reacts completely with one mole of hydroxide ions (OH-) from a base to form one mole of water under standard conditions.
(ii) Copper(II) chloride is soluble in water, so it can be made in solution by neutralizing an acid with a base and then crystallized. Lead(II) chloride is insoluble in cold water, so a neutralization in solution would simply precipitate it; it is therefore prepared by precipitation (double decomposition), not by neutralization.
(b)(i) Manufacture of trioxonitrate(V) acid (Ostwald process)
- Ammonia is mixed with excess air and passed over a platinum-rhodium gauze catalyst at about 900 °C:
\[4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O\]
- The nitrogen(II) oxide is cooled and oxidized by more air:
\[2NO + O_2 \rightarrow 2NO_2\]
- The nitrogen(IV) oxide is dissolved in water in the presence of excess air:
\[4NO_2 + O_2 + 2H_2O \rightarrow 4HNO_3\]
(ii) Sodium trioxonitrate(V) on strong heating gives sodium trioxonitrate(III) and oxygen:
\[2NaNO_3 \rightarrow 2NaNO_2 + O_2\]
(c)(i) \[3Mg + N_2 \rightarrow Mg_3N_2\]
(ii) Amount of Mg \(= \dfrac{3.0}{24} = 0.125\ \text{mol}\).
From the equation, 3 mol Mg give 1 mol Mg3N2, so
amount of Mg3N2 \(= \dfrac{0.125}{3} = 0.0417\ \text{mol} \approx 4.17 \times 10^{-2}\ \text{mol}\).
