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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)HCl(aq), as described by the chemical equationMnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)How much MnO2(s)MnO2(s) should be added to excess HCl(aq)HCl(aq) to obtain 115 mL Cl2(g)115 mL Cl2(g) at 25 °C and 805 Torr805 Torr?

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dsdrajlin

Answer:

4.98 × 10⁻³ mol

Explanation:

Given data for Cl₂

  • Volume (V): 115 mL = 0.115 L
  • Pressure (P): 1.06 atm

[tex]805torr.\frac{1atm}{760torr} =1.06atm[/tex]

  • Temperature (T): 25°C + 273.15 = 298 K

First, we will calculate the moles of Cl₂ using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.06 atm × 0.115 L / (0.0821 atm.L/mol.K) × 298 K

n = 4.98 × 10⁻³ mol

Let's consider the balanced equation.

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

The molar ratio of MnO₂ to Cl₂ is 1:1. The required moles of MnO₂ are 4.98 × 10⁻³ moles.

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