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).

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 =$25^0C=(25+273)K=298K$

$n=\frac{PV}{RT}$

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

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

According to stoichiometry:

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

Thus 0.0102 moles of chlorine is produced by = $\frac{1}{1}\times 0.0102=0.0102$ moles of $MnO_2$

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

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