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?

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kobenhavn kobenhavn · Answer 1 · 2020-05-07T07:19:29+02:00

Answer: 0.887 g of [tex]MnO_2[/tex] should be added to excess HCl(aq).

Explanation:

According to ideal gas equation:

[tex]PV=nRT[/tex]

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 =[tex]0.0821Latm/Kmol[/tex]

T =temperature =[tex]25^0C=(25+273)K=298K[/tex]

[tex]n=\frac{PV}{RT}[/tex]

[tex]n=\frac{1.06atm\times 0.235L}{0.0820 L atm/K mol\times 298K}=0.0102moles[/tex]

[tex]MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)[/tex]

According to stoichiometry:

1 mole of chlorine is produced by = 1 mole of [tex]MnO_2[/tex]

Thus 0.0102 moles of chlorine is produced by = [tex]\frac{1}{1}\times 0.0102=0.0102[/tex] moles of [tex]MnO_2[/tex]

Mass of [tex]MnO_2[/tex] =[tex]moles\times {\text {Molar mass}}=0.0102mol\times 87g/mol=0.887g[/tex]

0.887 g of [tex]MnO_2[/tex] should be added to excess HCl(aq).