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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

MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

How much MnO2(s) should be added to excess HCl(aq) to obtain 185 mL Cl2(g) at 25 °C and 745 Torr ?

mass of MnO2:

Sagot :

maacastrobr

Answer:

0.605 g

Explanation:

  • MnO₂(s) + 4HCl(aq) ⟶ MnCl₂(aq) + 2H₂O(l) + Cl₂(g)

First we calculate how many Cl₂ moles need to be produced, using the PV=nRT formula:

  • P = 745 Torr ⇒ 745 / 760 = 0.980 atm
  • V = 185 mL ⇒ 185 / 1000 = 0.185 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 25 °C ⇒ 25 + 273.16 = 298.16 K

Inputting the data:

  • 0.980 atm * 0.185 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
  • n = 0.00696 mol

Then we convert 0.00696 moles of Cl₂ to MnO₂ moles:

  • 0.00696 mol Cl₂ * [tex]\frac{1molMnO_2}{1molCl_2}[/tex] = 0.00696 mol MnO₂

Finally we convert 0.00696 moles of MnO₂ to grams, using its molar mass:

  • 0.00696 mol MnO₂ * 86.94 g/mol = 0.605 g
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