Amount of MnO₂ added: <u>0.6198 gr
</u>
<h3>Further explanation
</h3>
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
<h3>Learn more
</h3>
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