Question

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described by the chemical equation

MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

How much MnO2(s) should be added to excess HCl(aq) to obtain 385 mL Cl2(g) at 25 °C and 765 Torr ? mass of MnO 2

Answer 1

`m_(MnO_2)=1.378gMnO_2`

Step-by-step explanation:

Hello,

In this case, we first apply the ideal gas equation to compute the moles of produced chlorine:

`n_(Cl_2)=(PV)/(RT)=(765 Torr*(1atm)/(760Torr)*0.385L)/(0.082(atm*L)/(mol*K)*298.15K) =0.01585molCl_2`

Then, by considering the given reaction, applying the stoichiometry, that shows a 1 to 1 relationship between chlorine and manganese dioxide, we find:

`m_(MnO_2)=0.01585molCl_2*(1molMnO_2)/(1molCl_2) *(86.937gMnO_2)/(1molMnO_2) \\m_(MnO_2)=1.378gMnO_2`

Best regards.