Question

How many grams of potassium chlorate decompose to potassium chloride and $797$ mL of $128$ degrees Celsius and $747$ torr?
$2KCl{O_3}(s) \to 2KCl(s) + 3{O_2}(g)$

Answer 1

Hint: The moles of oxygen is calculated by the ideal gas equation. The balanced equation is used to calculate the moles of potassium chlorate. Consequently, the mass of potassium chlorate is calculated by the formula of the number of moles formula.

Complete step by step answer:
We know that the ideal gas equation is given by $PV = nRT$
where P is the pressure, n is the number of moles, V is the volume, R is the universal gas constant, T is the temperature.
The value of pressure of oxygen can be found out as follows:
$P = 747 \times \dfrac{1}{{760}} = 0.983\;atm$
The value of the volume is found out as follows:
$V = 797 \times \dfrac{{1L}}{{{{10}^3}mL}} = 0.797\;L$
The temperature is converted from degree celsius to kelvin temperature scale as
$T = 128 + 273 = 401\;K$
Now the moles of oxygen are found out by the ideal gas formula which is equal to the given pressure times volume divided by gas constant times temperature.
$ \Rightarrow n = \dfrac{{PV}}{{RT}} = \dfrac{{0.983 \times 0.797}}{{0.082057 \times 401}} = 0.0238$ mol
Now we will use the coefficient of the chemical equation to find the relative number of moles of potassium chlorate that reacted.
$ \Rightarrow 0.0238 \times \dfrac{2}{3} = 0.0159$ mol oxygen
Finally, we will use the molar mass of potassium chlorate which is $122.55\;g\;mo{l^{ - 1}}$ to find the number of grams that reacted
$ \Rightarrow 0.0159 \times \dfrac{{122.55}}{1} = 1.95$ g

Hence the grams of potassium chlorate which decomposes to potassium chloride is $1.95\;g$

Note: The balanced equation should be written for this kind of reaction. The mole can be found out by the ideal gas equation. The mole ratio of the compounds can be found out by the comparing of respective coefficients of the atoms.
The ideal gas is also called the general gas equation which explains the behaviour of a gas under various conditions.