Question f2548

1 Answer
Oct 8, 2016

Approx. 16% with respect to the starting mass of potassium nitrate.

Explanation:

We assume that (under fierce heating), $\text{potassium nitrate}$ decomposes to give stoichiometric $\text{potassium nitrite}$ and $\text{dioxygen gas}$:

$K N {O}_{3} \left(s\right) + \Delta \rightarrow K N {O}_{2} \left(s\right) + \frac{1}{2} {O}_{2} \left(g\right) \uparrow$

And thus $1$ $m o l$ of $\text{potassium nitrate}$ loses $16 \cdot g$ of oxygen gas upon decomposition. With respect to $K N {O}_{3}$, this represents a mass loss of:

(16*g)/(101.17*g)xx100% $\cong$ 16%#