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 ?

0.86 g

Explanation:

Given:  

Pressure = 725 torr

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

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,

1 mole of chlorine is formed from 1 mole of

Thus, moles of = 0.0099 moles

Molar mass of = 86.9368 g/mol

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

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