# What mass of "potassium chlorate" is required to produce a 2.24*L volume of dioxygen gas at "NTP"?

Aug 7, 2017

We need (i) a stoichiometric equation, and (ii)......

#### Explanation:

and........(ii) we need the molar volume at $N T P$, which I think is $24.04 \cdot L \cdot m o {l}^{-} 1$.......

Now as regards (i)......

$K C l {O}_{3} \left(s\right) + \Delta \stackrel{M n {O}_{2}}{\rightarrow} K C l \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \uparrow$

The reaction generally works well if you add some $\text{manganese (IV) oxide}$ as an oxygen transfer reagent.

And so moles of dioxygen gas......$= \frac{2.24 \cdot L}{24.04 \cdot L \cdot m o {l}^{-} 1}$

$= 0.0932 \cdot m o l$

And so the molar quantity of $K C l {O}_{3}$ is......

$\frac{2}{3} \times 0.0932 \cdot m o l \times 122.55 \cdot g \cdot m o {l}^{-} 1 = 7.61 \cdot g$.

Look at the question again, are you sure that it did not ask for $\text{STP}$? The different definitions and standards vary across each syllabus; there are different definitions in yurope, Blighty, and the States. So you need to look up your definition of $\text{NTP}$.