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

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23-03-2020
Chemistry

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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

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sebassandin sebassandin · Answer 1 · 2020-03-27T01:19:37+01:00

Answer:

[tex]m_{MnO_2}=1.378gMnO_2[/tex]

Explanation:

Hello,

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

[tex]n_{Cl_2}=\frac{PV}{RT}=\frac{765 Torr*\frac{1atm}{760Torr}*0.385L}{0.082\frac{atm*L}{mol*K}*298.15K} =0.01585molCl_2[/tex]

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

[tex]m_{MnO_2}=0.01585molCl_2*\frac{1molMnO_2}{1molCl_2} *\frac{86.937gMnO_2}{1molMnO_2} \\m_{MnO_2}=1.378gMnO_2[/tex]

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