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

Answer:

0.88 g

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:

$805torr\times 0.235L=n\times 62.3637\text{ torrHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{805\times 0.235}{62.3637\times 298}=0.01017\ mol$

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,

$Mass=Moles\times Molar\ mass$

Applying values, we get that:-

$Mass=0.01017moles \times 86.93685\ g/mol=0.88\ g$

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.