Chemistry, 13.01.2020 01:31, natetheman7740

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?

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Answers

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

answered: anna2894

Amount of MnO₂ added: 0.6198 gr

Further explanation

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

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:

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

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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Hbr dissociates to form h+ and br-ca(oh)2 dissociates to form ca2+ and 2oh-the h+ and oh- react to form waterassuming the reactants are dissolved in water, the net reaction is h+ + oh- —> h2oif you’re asked to show all components of the reaction, the complete reaction is 2hbr + ca(oh)2 —> cabr2 + 2h2o

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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric ac...