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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 Ā°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 Ā°C and 715 Torr?

Respuesta :

Ā  As we knwĀ 
Ā P = 715/760 = 0.94078atmĀ 
v=185ml=0.185l
n = ? moles have to find
R = 0.0821 L atm/K/moleĀ 
T = 25 + 273 = 298 KĀ 
as
Ā PV = nRTĀ 
putting values so
0.94078*0.185=nĀ 0.0821*298Ā 
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
soĀ 0.00711191 of chlorine must have come fromĀ 0.00711191 moles of MnO2
Ā 1 mole of MnO2 = 86.94 g/mole
soĀ Ā 0.00711191 moles of MnO2== 86.94*Ā 0.00711191
=0.61830946
hope it helps
ardni313

Amount of MnOā‚‚ added: 0.6198 gr

Further explanation

In general, the gas equation can be written

[tex] \large {\boxed {\bold {PV = nRT}}} [/tex]

where

P = pressure, atm, N / mĀ²

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 = mĀ³)

T = temperature, Kelvin

n = N / No

n = mole

No = Avogadro number (6.02.10Ā²Ā³)

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 Ā°C = 25 + 273 = 298 K

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

[tex]\rm n=\dfrac{PV}{RT}\\\\n=\frac{0.941. 0.185}{0.082.298}\\\\n=0.007124\\\\n=7,124\times 10^{-3}[/tex]

From the reaction:

4HCL + MnOā‚‚ ---> MnClā‚‚ + Clā‚‚ + 2Hā‚‚O

because HCl as excess reactant, MnOā‚‚ as limiting reactant so that the mole ratio is the same as Clā‚‚

mol MnOā‚‚ = mol Clā‚‚ = 7,124. 10ā»Ā³

molar mass of MnOā‚‚ = 87 Ā g/mol

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

gram MnOā‚‚ = 7,124. 10ā»Ā³ x 87

gram MnOā‚‚ = 0.6198

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