Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, hcl(aq), as described by the chemical equation.
how much mno2(s) should be added to excess hcl(aq) to obtain 185 ml of cl2(g) at 25 °c and 715 torr?
chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, hcl(aq), as described by the chemical equation.
how much mno2(s) should be added to excess hcl(aq) to obtain 185 ml of cl2(g) at 25 °c and 715 torr?

  As we knw 
 P = 715/760 = 0.94078atm 
v=185ml=0.185l
n = ? moles have to find
R = 0.0821 L atm/K/mole 
T = 25 + 273 = 298 K 
as
 PV = nRT 
putting values so
0.94078*0.185=n 0.0821*298 
24.466n=0.1740443
n=0.174/24.466
n=0.00711191 nb of moles of cl2
as 1 mole of Cl2 were obtained from 1 mole of MnO2
so 0.00711191 of chlorine must have come from 0.00711191 moles of MnO2
 1 mole of MnO2 = 86.94 g/mole
so  0.00711191 moles of MnO2== 86.94* 0.00711191
=0.61830946
hope it helps

Amount of MnO₂ added: 0.6198 gr

Further explanation

In general, the gas equation can be written

where

P = pressure, atm, N / m²

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K (P = atm, v = liter), or 8.314 J / mol K (P = Pa or N / m2, v = m³)

T = temperature, Kelvin

n = N / No

n = mole

No = Avogadro number (6.02.10²³)

n = m / m

m = mass

M = relative molecular mass

Assuming an ideal gas then

1 torr = 0.00131579 atm

P = 715 torr = 0.941 atm

V = 185 ml = 0.185 L

T = 25 °C = 25 + 273 = 298 K

so the number of moles of Cl₂ formed can be found from the ideal gas equation:

From the reaction:

4HCL + MnO₂ ---> MnCl₂ + Cl₂ + 2H₂O

because HCl as excess reactant, MnO₂ as limiting reactant so that the mole ratio is the same as Cl₂

mol MnO₂ = mol Cl₂ = 7,124. 10⁻³

molar mass of MnO₂ = 87  g/mol

then the number of grams of MnO₂ to be added: mol x molar mass

gram MnO₂ = 7,124. 10⁻³ x 87

gram MnO₂ = 0.6198

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