Question

What volume of dioxygen gas would be generated by complete decomposition of a `10*g` mass of `KClO_3`, under a pressure of `750*mm*Hg`?

Answer 1

Well, we need a stoichiometric equation......and I get under `2*L`.

Explanation:

`KClO_3stackrel(MnO_2)rarr3/2O_2(g) + KCl`

Some `Mn(IV)` salt is usually added to catalyze the reaction.

`"Moles of potassium chlorate"=(10*g)/(122.55*g*mol^-1)`

`=0.0816*mol`

And thus we gets `3/2xx0.0816*mol=0.122*mol` `"dioxygen gas"`.

And then we simply solve the Ideal Gas equation.....

`V=(nRT)/P=(0.0816*molxx0.0821*(L*atm)/(K*mol)xx291*K)/((750*mm*Hg)/(760*mm*Hg*atm^-1))`

`=??*L`

Note the pressure measurement. A mercury column is a VERY convenient means to measure gas pressure in that we know that `1*atm` will support a column of mercury that is `760*mm` high. These days mercury has all but disappeared from laboratories because of perceived safety concerns. (I keep my manometers under lock and key, away from nosy safety inspectors).