Question

In the reaction,
$2KCl{O_3}(s)\, \to \,2KCl(s)\, + \,3{O_2}(g)$
How many liters of oxygen would be produced from $231\,g$ of potassium chlorate?

Answer 1

The standard temperature and pressure (STP) conditions are defined at $100{\text{ }}kPa$ or $1{\text{ }}bar$ of pressure and $0^\circ C$ of temperature. Under these conditions for pressure and temperature, the molar volume of a gas at STP is defined as one mole of any ideal gas occupies $22.7{\text{ }}Litres$ .
So, to find the volume of oxygen gas at Standard temperature and pressure, we need to know how many moles of oxygen are produced by this reaction.

Complete step-by-step answer: Since in this reaction we need to assume that it is happening in standard temperature and pressure which is denoted as STP.
Let’s write the balanced chemical equation for this decomposition reaction;
$2KCl{O_3}(s)\,\,\xrightarrow{{heat}}\,\,2KCl(s)\, + \,3{O_2}(g)\, \uparrow$
We are able to observe that we have a $2:3$ mole ratio between potassium chlorate $(KCl{O_3})$ and oxygen gas$({O_2})$ .
It explains to us that this reaction will always produce $\dfrac{3}{2}$ times more moles of oxygen gas than the number of moles of potassium chlorate which underwent the decomposition reaction.
Let’s use potassium chlorate's molar mass to determine how many moles we have in that $231\,g$ sample
As we know, the number of moles will be equal to the given mass divided by the molecular mass a compound.
Hence, the formula will be as follows,
$n\,\, = \,\,\dfrac{{mass}}{{molar\,mass}}$
Where,
$n\, =$ the amount in moles $(mol)$
Mass will be in the terms of $\;(g)$
Molar mass will be in the terms of $\;(g/mol)$
So, we need to calculate the moles of $KCl{O_3}$ .
first, we need to calculate the molar mass of $KCl{O_3}$and then we need to calculate the moles.
Let’s calculate the molar mass of this compound $KCl{O_3}$ ;
The molar mass of potassium $= \,39\,g/mol$
We have one potassium so the overall molar mass for potassium will be;
$= \,39\, \times \,1\, = \,39\,g/mol$
The molar mass of chlorine $= \,35.5\,g/mol$
We have one chlorine so the overall molar mass for chlorine will be;
$= \,35.5\, \times \,1\, = \,35.5\,g/mol$
The molar mass of oxygen $= \,16\,g/mol$
We have three oxygen so the overall molar mass for oxygen will be;
$= \,16\, \times \,3\, = \,48\,g/mol$
Now the total molar mass of $KCl{O_3}$ will be;
$= \,39\, + \,35.5\, + 48$
$= \,122.5\,g/mol$
Now, the given mass of KCl{O_3}$$$$ = \,231\,g
Let’s substitute in the values;
$n\,\, = \,\,\dfrac{{\,231\,g}}{{122.5\,g/mol}}\,$
$= \,1.885\,moles$ of $KCl{O_3}$
Now, let’s find how many moles of oxygen would be produced from potassium chlorate;
$= \,1.885\,moles\,of\,KCl{O_3}\, \times \,\dfrac{{3\,moles\,{O_2}}}{{2\,moles\,of\,KCl{O_3}}}$
$= \,2.827\,moles\,of\,{O_2}$
Let’s find how many liters of oxygen would be produced;
So, we need to multiply with 22.7 L with $2.827\,moles\,of\,{O_2}$
$= \,2.827\, \times \,22.7$

Note: Standard Temperature and Pressure. Standard temperature is equal to $0^\circ C$ , which is $273.15{\text{ }}K$ . Standard Pressure is $1{\text{ }}Atm$ , $100{\text{ }}kPa$ or $760{\text{ }}mmHg$ or $torr$ . STP is the "standard" condition often used for measuring gas density and volume.