Question

100ml of 0.2mole of HCl is treated with excess MnO2.what is the volume of Cl2 obtained after tasting?

Answer 1

Answer:

0.112L Cl2

Explanation:

4HCl + MnO2= MnCl2 + Cl2 +2H2O

1000ml 4 M HCl= 22.4 L Cl2

100ml 0.2 M HCl= {22.4×0.2×100}/(1000×4)

=0.112L Cl2

Answer 2

Answer:

$0.112 L$

Explanation:

Given: Here we have $100 ml$  $0.2$ a mole of $HCl$ is treated with excess $MnO_{2}$.

To find: We have to find the volume  $Cl_{2}$ obtained after tasting.

Concept: The total number of moles of solute in a given solution's molarity is expressed as moles of solute per liter of solution. As opposed to mass, which fluctuates with changes in the system's physical circumstances, the volume of a solution depends on changes in the system's physical conditions, such as pressure and temperature. M, sometimes known as a molar, stands for molarity. When one gramme of solute dissolves in one litre of solution, the solution has a molarity of one. Since the solvent and solute combine to form a solution in a solution, the total volume of the solution is measured.

The volume of solvent required to dissolve the provided solute is multiplied by the ratio of the moles of the solute whose molarity has to be computed.

$M=\frac{n}{v}$

The solution's molality, M, needs to be computed.

n is the solute's molecular weight in moles.

V represents the solution's volume in liters.

Step 1

$4HCl+MnO_{2}$ ⇒ $MnCl_{2} +Cl_{2} +2H_{2} O$

As we know that $1000ml$ $4M HCl$ = $22.4L Cl_{2}$

So, $100ml$ $0.2 M HCl =$ $\frac{(22.4\cdot0.2\cdot100)}{(1000\cdot4)}$

                                 $= 0.122 l$

Hence, the volume will be $0.122 l$.

#SPJ2