Question

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid. HCl(aq), as described by the chemical equationMnO2(s) + 4HCl(aq) --> MnCl2(aq) + 2H2O(l) + Cl2(g)How much MnO2(s) should be added to excess HCl (aq) to obtain 235 mL of CL2(g) at 25 degrees C and 805 Torr?

Answer 1

0.88 g

Step-by-step explanation:

Using ideal gas equation to calculate the moles of chlorine gas produced as:-

`PV=nRT`

where,

P = pressure of the gas = 805 Torr

V = Volume of the gas = 235 mL = 0.235 L

T = Temperature of the gas =
`25^oC=[25+273]K=298K`

R = Gas constant =
`62.3637\text{torr}mol^(-1)K^(-1)`

n = number of moles of chlorine gas = ?

Putting values in above equation, we get:

`805torr* 0.235L=n* 62.3637\text{ torrHg }mol^(-1)K^(-1)* 298K\\\\n=(805* 0.235)/(62.3637* 298)=0.01017\ mol`

According to the reaction:-

`MnO_2+4HCl\rightarrow MnCl_2+2H_2O+Cl_2`

1 mole of chlorine gas is produced when 1 mole of manganese dioxide undergoes reaction.

So,

0.01017 mole of chlorine gas is produced when 0.01017 mole of manganese dioxide undergoes reaction.

Moles of
`MnO_2` = 0.01017 moles

Molar mass of
`MnO_2` = 86.93685 g/mol

So,

`Mass=Moles* Molar\ mass`

Applying values, we get that:-

`Mass=0.01017moles * 86.93685\ g/mol=0.88\ g`

0.88 g of
`MnO_2(s)` should be added to excess HCl (aq) to obtain 235 mL of
`Cl_2(g)` at 25 degrees C and 805 Torr.