Question

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 235 mL Cl2(g)235 mL Cl2(g) at 25 °C and 805 Torr805 Torr?

Answer 1

Answer: 0.887 g of
`MnO_2` should be added to excess HCl(aq).

Step-by-step explanation:

According to ideal gas equation:

`PV=nRT`

P = pressure of gas = 805 torr = 1.06 atm (760torr=1atm)

V = Volume of gas = 235 ml = 0.235 L

n = number of moles = ?

R = gas constant =
`0.0821Latm/Kmol`

T =temperature =
`25^0C=(25+273)K=298K`

`n=(PV)/(RT)`

`n=(1.06atm* 0.235L)/(0.0820 L atm/K mol* 298K)=0.0102moles`

`MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)`

According to stoichiometry:

1 mole of chlorine is produced by = 1 mole of
`MnO_2`

Thus 0.0102 moles of chlorine is produced by =
`(1)/(1)* 0.0102=0.0102` moles of
`MnO_2`

Mass of
`MnO_2` =
`moles* {\text {Molar mass}}=0.0102mol* 87g/mol=0.887g`

0.887 g of
`MnO_2` should be added to excess HCl(aq).