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)

Calculate the weight of MnO_(2) and the volume of HCl of specific grav

1.	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)
Answer» Solution :(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`