Question

Calculate the mass of potassium chloride and the volume of oxygen evolved at ${27^ \circ }C$ and $750mm$ of mercury, when $24.5g$ of potassium chlorate is heated.
A.$24.5g,6.48L$
B.$19.2g,8.40L$
C.$14.9g,7.48L$
D.$34.53g,9.84L$

Answer 1

In order to calculate the mass of potassium chloride and the volume of oxygen we have to calculate the molecular mass of potassium chlorate so that we can calculate the number of moles of potassium chloride and after that we will apply the relation between molecular mass and number of moles of compound we can calculate the mass.

Complete step by step answer:
A.First we will calculate the molecular weight of $KCl{O_3}$ ,
Atomic mass of potassium $= 39$ , atomic mass of chlorine $= 35.5$ ,atomic mass of oxygen $= 16$ Molecular weight of $KCl{O_3} =$ atomic mass of potassium $+$ atomic mass of chlorine $+$ atomic mass of oxygen $\times 3$
Substituting the values we get,
Molecular weight of $KCl{O_3} = 39 + 35.5 + 16 \times 3$
Molecular weight of $KCl{O_3} = 74.5 + 48$
Molecular weight of $KCl{O_3} = 122.5$
B.Given data:
Molecular weight of $KCl{O_3} = 122.5$
Weight of $KCl{O_3} =$ $24.5$
Moles of $KCl{O_3} = \dfrac{W}{{MW}}$
Substituting the value from the given data we get,
Moles of $KCl{O_3} = \dfrac{{24.5}}{{122.5}}$
Moles of $KCl{O_3} = 0.2$
C.GIVEN DATA:
Moles of $KCl = 0.2$
Atomic mass of potassium $= 39$ , atomic mass of chlorine
Molecular weight of $KCl =$ atomic mass of potassium $+$ atomic mass of chlorine
Substituting the values we get,
Molecular weight of $KCl = 39 + 35.5$
Molecular weight of $KCl = 74.5$
Moles of $KCl = \dfrac{W}{{MW}}$
Weight of $KCl = MW \times$ number of moles
Substituting the values we get
Therefore, weight of $KCl = 0.2 \times 74$
Weight of $KCl = 14.9g$
D.In this to find out the volume of oxygen we will use the ideal gas equation.
First we will see what is ideal gas and then its equation:
Ideal gas is defined as the gas that obeys all the gas laws at all conditions of temperatures and pressure.
Ideal gas law states that the pressure, temperature and volume are inter related to each other by the formula given as follows:
$PV = nRT$
Where,
$P =$ Pressure of gas
$V =$ Volume of gas
$n =$ number of moles of gas.
$R =$ Gas constant
$T =$ temperature
To find volume of gas we will use the formula of ideal gas by rearranging it as follows:
$PV = nRT$
$V = \dfrac{{nRT}}{P}$
The reaction is given as follows:
$2KCl{O_3} \to 2KCl + 3{O_2}$
Given data:
Temperature $T = {27^ \circ }C$
$T = 27 + 273$
$T = 300K$
Pressure $P = 750mmHg$
$P = \dfrac{{750}}{{760}}$
$P = 0.98atm$
Number of moles $n = 0.2$
$R = 8.314$

Using the formula: $V = \dfrac{{nRT}}{P}$
Substituting the values from the given data we get,
$V = \dfrac{3}{2} \times \dfrac{{0.2 \times 0.0821 \times 300}}{{0.98}}$
$V = 7.48L$
So, the correct answer is option c) $14.9g,7.48L$ .
($\dfrac{3}{2}$ is that $3$ moles of oxygen produces $2$ moles of potassium chloride)

Note:
-Ideal gas law defined the relationship between pressure, volume, number of moles of gas and temperature. This law combines Boyle's law and Charles law. Gas constant has two values that is $8.314$ and $0.0821$.The use of these depends upon the type of units which want an answer in. If we want an answer in L.atm/K/mol then consider $0.0821$ value as gas constant and if we want answer in J/K/mol then consider $8.314$ value as gas constant.