JEE Advance - Chemistry (2017 - Paper 2 Offline - No. 12)
Explanation
(a) Conc. HNO3 makes iron passive. Cold relatively concentrated HNO3 will react with Fe.
$$\mathop {Fe}\limits_{Iron} + \mathop {6HN{O_3}}\limits_{Nitric\,acid} \to \mathop {Fe{{(N{O_3})}_3}}\limits_{Iron\,nitrate} + \mathop {3N{O_2}}\limits_{Nitrogen\,dioxide} + \mathop {3{H_2}O}\limits_{Water} $$
(b) $$\mathop {Cu}\limits_{Copper} + \mathop {4HN{O_3}(conc.)}\limits_{Nitric\,acid} \to \mathop {Cu{{(N{O_3})}_2}}\limits_{Copper\,nitrate} + \mathop {2N{O_2}}\limits_{Nitrogen\,dioxide} + \mathop {2{H_2}O}\limits_{Water} $$
(c) $$\mathop {Zn}\limits_{Zinc} + \mathop {2NaO{H_{(aq)}}}\limits_{Sodium\,hydroxide} \to \mathop {N{a_2}Zn{O_2}}\limits_{Sodium\,zincate} + \mathop {{H_2}}\limits_{Hydrogen} $$
(d) $$\mathop {4Au}\limits_{Gold} + \mathop {8NaCN}\limits_{Sodium\,cyanide} + \mathop {{O_2}}\limits_{Oxygen} + \mathop {2{H_2}O}\limits_{Water} \to \mathop {4Na[Au{{(CN)}_2}]}\limits_{Sodium\,dicyanoaurate} + \mathop {4NaOH}\limits_{Sodium\,hydroxide} $$
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