(a) Explain the differences in the reactions of zinc with dilute trioxonitrate (V) acid and zinc with dilute hydrochloric acid.
(b) Write equations to illustrate how ammonia gas can be converted into trioxonitrate (V) acid.
(c) Calculate the mass of sodium trioxonitrate (V) produced when 30.0g of pure sodium hydroxide reacts with 100cm\(^3\) of 1.00 M trioxonitrate (V) acid. (H =1, N = 14, 0 = 16, Na = 23)
(d) Write the equations for the decomposition by heat of:
(i) sodium trioxonitrate(V);
(ii) copper (II) trioxonitrate (V);
(iii) mercury (II) trioxonitrate (V);
Explanation
(a) Zinc reacts with dilute hydrochloric acid to produce hydrogen as but with trioxonitrate (VI) acid. the hydrogen initially produced is at once oxidized by the powerful oxidizing ability of the trioxonitrate (V) acid to give water and the reduction products of the acid.
(b) Step I 4NH\(_{3(g)}\) + SO\(_{2(g)}\) --> 4NO\(_{(g)}\) + 6H\(_2\)O\(_{(l)}\)
Step II 2NO\(_{(g)}\) + O\(_{2(g)}\) —> 2NO\(_{2(g)}\)
Step III 2H\(_2\)O\(_{(l)}\) + 4NO\(_{2(g)}\) + O\(_{2(g)}\) —> 4HNO\(_{3(aq)}\)
(c) NaOH(\(_{(aq)}\) + HNO\(_{3(aq)}\) —> NaNO\(_{3(aq)}\) + H\(_2\)O\(_{(l)}\)
30g NaOH x \(\frac{1 mole}{g}\)
Since mole = concentration x volume
Mole HNO\(_3\) = 1.00 mole x 100dm\(^3\) = 0.1 mole dm\(^3\)
From the balanced neutralization reaction,
\(\frac{NaOH}{HNO_ 3}\) = \(\frac{1}{1}\)
i.e mole NaOH = mole HNO\(_3\). The limiting reagent in the above rxn, is 0.1 mole HNO\(_3\)
0.1 mole HNO\(_3\) will produce 85 x 0.1 = 8.5g NaNO\(_3\)
(d)(i) 2NaNO\(_{3(g)}\) \(\to\) 2NaNO\(_{2(g)}\) + O\(_{2(g)}\)
(ii) 2Cu(NO\(_3\))\(_{2(g)}\) \(\to\) 2CuO\(_{(g)}\) + 4NO\(_2\) + O\(_{2(g)}\)
(iii) Hg(NO\(_3\))\(_{2(g)}\) -> Hg\(_{(l)}\) + 2NO\(_{2(g)}\) + O\(_{2(g)}\)