Question

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described by the chemical equation MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) How much MnO2(s) should be added to excess HCl(aq) to obtain 255 mL Cl2(g) at 25 °C and 725 Torr ?

Answer 1

Answer:

0.86 g

Explanation:

Given:  

Pressure = 725 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,  

Pressure = 725 / 760 atm = 0.9539 atm

Volume = 225 mL = 0.225 L (1 mL = 0.001 L)

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9539 atm × 0.255 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 0.0099 moles

From the reaction,

$MnO_2_{(s)}+4HCl_{(aq)}\rightarrow MnCl_2_{(aq)}+2H_2O_{(l)}+Cl_2_{(g)}$

1 mole of chlorine is formed from 1 mole of $MnO_2$

Thus, moles of $MnO_2$ = 0.0099 moles

Molar mass of $MnO_2$ = 86.9368 g/mol

Mass = Moles*Molar mass = 0.0099 moles * 86.9368 g/mol = 0.86 g