If `1 g` of `HCl` and `1 g` of `MnO_(2)` heated together the maximum weight of `Cl_(2)` gas evolved will be :
`[MnO_(2) + 4HCl rarr MnCl_(2) + Cl_(2) + 2H_(2)O]` :

If 1 gm of HCl and 1 gm of MnO_2 are heated together, the maximum weight of Cl_2 gas evolved will be:

What is the maximum weight of Cl_(2) obtained by the action of 1g HCl on 1g MnO_(2) ?

In the reaction MnO_(2)+4HCl to MnCl_(2)+2H_(2)O+Cl_(2)

Calculated the weight of MnO_2 required to produce 1.78 litre of Cl_2 at STP by following reaction MnO_2 + 4 HCl rarr MnCl_2 + Cl_2 + 2H_2O

Balance the chemical equation by the oxidation number method. (viii) MnO_(2) + HCl to MnCl_(2) + Cl_(2) + H_(2)O

Balance the following equation : MnO2 + HCl → MnCl2 + Cl2 + H2O

Balance the following equation : MnO_2 + HCl to MnCl_2 + H_2O + Cl_2 .

Calcualte the volume of Cl_(2) gas (in ml) liberated at 1 atm 273 K when 1.74 gm MnO_(2) reacts with 2.19 gm HCl according to the following with % yeild 40. MnO_(2) + 4HCl rightarrow MnCl_(2) + Cl_(2) + 2H_(2)O

Balance the equation: MnO_4^- + H + +Cl rarr Mn^(1+) +Cl_2 +H_2O