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\(_{2(g)}\)

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