Question

What volume in dm3 of KCl is obtained in the following equation ??

Answer 1

The method is:

1) Calculate the grams of KCl produced.

   - you need to know the starting quantity of reactant KClO3, which usually is in grams.
   - convert the grams to moles, by dividing by the molar mass of the KClO3
   - use the ratio 2 moles of KClO3 produces 2 moles of KCl
   - convert the moles of KCl to grams, by multiplying by the molar mass of KCl

2) Given that the product is in solid state (which is weird), you will need to use the apparent density of the KCl, which is a datum that you have to search in tables.

It would be more logical to ask for the volume of O2 which is in gas state. If this were the case, you should know, the temperature, and the pressure, to use solve for V from the ideal gas equation: pV = nRT => V = nRT/p

n: number of moles produced of the gas (using the ratio 2 moles of KClO3 produces 2 moles of O2)

T: temperature in kelvin
p: pressure
R: Universal constant of gases
V: volume of O2.

Answer 2

Answer: Volume of KCl is $0.0752dm^3$

Explanation: For the given reaction,

$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

We can see that 2 moles of $KClO_3$ is giving 2 moles of KCl. So, the number of moles of KCl will be 2.

Mass of KCl can be calculated by,

$\text{Number of Moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of KCl = 74.5 g/mol.

Number of Moles = 2

$\text{Given mass}= Moles\times \text{Molar Mass}$

Given mass = 149g

Now, to calculate the Volume of KCl, we will use the density formula,

$\rho =\frac{Mass}{Volume}$

$\text{Specific density of KCl}(\rho)=1.98g/cm^{3}$

Mass = 149g (Calculated above)

Putting the values in density formula, we get

$1.98g/cm^3=\frac{149g}{Volume}$

$\text{Volume of KCl}=75.252cm^3$

Conversion Factor: $1cm^3=10^{-3}dm^3$

Volume of KCl in $dm^3=0.0752dm^3$