Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)HCl(aq), as described by the chemical equation MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) How much MnO2(s)MnO2(s) should be added to excess HCl(aq)HCl(aq) to obtain 115 mL Cl2(g)115 mL Cl2(g) at 25 °C and 805 Torr805 Torr?

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Asked Oct 15, 2019
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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)HCl(aq), as described by the chemical equation

 

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

 

How much MnO2(s)MnO2(s) should be added to excess HCl(aq)HCl(aq) to obtain 115 mL Cl2(g)115 mL Cl2(g) at 25 °C and 805 Torr805 Torr?

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Expert Answer

Step 1

Given values

Volume of chlorine to be obtained = 115ml

Temperature = 25°C

                        = (25+ 273.15)K = 298.15K

Pressure = 805 torr

Step 2

 

Volume of Cl2= 115ml

Since 1ml = 0.001L

Therefore, volume of Cl2= 0.115L

Step 3

According to ideal ...

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PV=nRT to calculate the moles, rearranging the equation PV RT R 62.36 torr mol'K-1 L 805torr x 0.115L Bel 62.36torr mol'K'Lx 298K 0.00498 moles

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