Question

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

• A. $N_2$
• B. $O_2$
• C. $NO_2$
• D. $Na_2O$

Answer 1

Answer: (D)

$Na_2O$

Answer 2

$NaNO_3$ when heated, it decomposes in two stages as:
$NaNO_3 \xrightarrow {T < 500^\circ C} NaNO_2 + \frac{1}{2} O_2$
$NaNO_2 \xrightarrow {T > 800^\circ C} Na_2O + N_2 + O_2$