Question

Sodium nitrate decomposes above $\approx ;800^{\circ }C$ to give

• A. $N_2$
• B. $O_2$
• C. $N{O}_2$
• D. $N{a}_{2}O$

Answer 1

Answer: (A), (B), (D)

$NaN{O}_3$ when heated, it decomposes in two stages as:
$NaN{O}_3\stackrel{T<500^{\circ }C}{\to }NaN{O}_2+\frac{1}{2}{O}_2$
$NaN{O}_2\stackrel{T>800^{\circ }C}{\to }N{a}_{2}O+N_2+O_2$