Question

Find out the volume of $Cl_{2}$ at STP produced by action of 100 cc of 0.2M HCl on excess of $MnO_{2}$.

Answer 1

The reaction between ${ MnO }_{ 2 }$ and HCl is as follows

${ MnO }_{ 2 }\quad +\quad 4HCl\quad \rightarrow \quad Mn{ Cl }_{ 2 }\quad +\quad 2{ H }_{ 2 }O\quad +\quad { Cl }_{ 2 }$

Molar mass of HCl = 36.5

$4HCl\quad =\quad 36.5\quad \times \quad 4\quad =\quad 146$

The volume of $Mn{ Cl }_{ 2 }\quad at\quad STP = 22400 cc$

$HCl\quad present\quad in\quad 100\quad cc\quad of\quad 0.2M\quad HCl\quad =\quad 0.02\quad millimoles\quad \times \quad 36.5\quad =\quad 0.73 g$

$146 g\quad of\quad HCl\quad produces\quad =\quad 22400\quad cc\quad of\quad { Cl }_{ 2 }\quad at\quad STP$

$0.73 g\quad of\quad HCl\quad produces\quad =\quad \frac { 22400\quad \times \quad 0.73 }{ 146 } \quad =\quad 112\quad cc\quad of\quad { Cl }_{ 2 }$