Calculate the weight of `MnO_(2)` and the volume of HCl of specific gravirty 1.2 g g `mL^(-1)` and `5%` by weight needed to produce 1.12 L of `Cl_(2)` at STP by the reaction `MnO_(2)+4"HCl"toMnCl_(2)+3H_(2)O+Cl_(2)`

Explanation

(a). `N_(HCl)=(% by weightxx10xxd)/(Ew of HCl)`
`=(5xx10xx1.2)/(36.5)=1.64`
Now, m" Eq of "`MnO_(2)-=m" Eq of "HCl`
`-=m" Eq of "Cl_(2) formed`.
`-=(1.12)/(11.2)xx10^(3)`
`=100[{:("Ew of "Cl_(2)=(M)/(2)),(1 " Eq of "Cl_(2)=11.2L):}]`
(b). Volume of HCl used:
`NxxV=100`
`1.64xxV=100`
`thereforeV_(HCl)to60.97mL`
Bacause HCl is also used to give `MnCl_(2)` thus volume used is double that required for the reduction of `MnO_(2)`.
`V_(HCl)=2xx60.97=121.94mL`
(c). Also, m" Eq of "`MnO_(2)=m" Eq of "HCl=100`
`(W)/((87)/(2))xx10^(3)=100(Ew of MnO_(2)=(55+32)/(2))`
`thereforeW_(MnO_(2))=4.35g`