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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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

Answer from: 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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Answer from: Quest

Cis the answer ur welcome if i get it wrong u don't have to me

Answer from: Quest

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. $hbr$ completely dissociates to $h^+$ and $br-$ ions. a is incorrect because $h_2s$ does not fully ionize to $h^+$ and $hs^-$ ions and some remain as the compound $h_2s$ . c and d are incorrect because, $li_2o$ and $libr$ do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise, if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both $li_2o$ and $libr$ have a complete octet so can not accept any electrons which disqualifies them as acids.

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