Q.Which metal nitrate gives colourless gas on thermal decomposition?

Dilute nitric acid oxidises zinc to Zn(NO$_3$)$_2$ while being reduced to nitrous oxide (N$_2$O); honouring the printed 5 H$_2$O, the completed equation is 4 Zn + 10 HNO$_3$ (dilute) → 4 Zn(NO$_3$)$_2$ + N$_2$O + 5 H$_2$O.

Zinc is a good reducing agent. With dilute $HNO_3$ the nitrogen is reduced from $+5$ (in $HNO_3$) to $+1$ (in $N_2O$), while each zinc atom is oxidised, $Zn \to Zn^{2+} + 2e^-$. The printed equation already fixes the water term as $5\,H_2O$, so the two blanks are simply the two products, which we find by balancing atoms:

• Zinc: the 4 Zn on the left give 4 $Zn(NO_3)_2$ on the right.
• Nitrogen: the 10 N (from 10 $HNO_3$) split as 8 N tied up as nitrate spectator ions in the 4 $Zn(NO_3)_2$, and 2 N reduced into one $N_2O$.
• Hydrogen: all 10 H (from 10 $HNO_3$) appear in the printed $5\,H_2O$.
• Oxygen: 30 O on the left $=$ 24 O in 4 $Zn(NO_3)_2$ $+$ 1 O in $N_2O$ $+$ 5 O in $5\,H_2O$ $=$ 30 O. ✓ …

Nitric acid = HNO3.

Nitric acid is a strong oxoacid of nitrogen with formula HNO3 (nitrogen oxidation state +5). The distractors are other nitrogen species: HNO2 is nitrous acid (N in +3) …

NaNO₃ → NaNO₂ + ½O₂ (colourless); heavy-metal nitrates give brown NO₂.

Nitrates of very electropositive metals (Na, K) decompose only to the nitrite and oxygen:

2NaNO₃ → 2NaNO₂ + O₂ (O₂ colourless).

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