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 equation MnO2(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 265 mL Cl2(g)265 mL Cl2(g) at 25 °C and 785 Torr785 Torr ?

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dsdrajlin dsdrajlin · Answer 1 · 2021-05-04T02:29:11+02:00

Answer:

0.0112 mol

Explanation:

Step 1: Write the balanced equation

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

Step 2: Convert 785 Torr to atm

We will use the conversion factor 1 atm = 760 Torr.

785 Torr × 1 atm/760 Torr = 1.03 atm

Step 3: Convert 25°C to K

We will use the following expression.

K = °C + 273.15 = 25°C + 273.15 = 298 K

Step 4: Calculate the moles of Cl₂(g)

265 mL (0.265 L) of Cl₂(g) are at 1.03 atm and 298 K. We can calculate the number of moles using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.03 atm × 0.265 L / (0.0821 atm.L/mol.K) × 298 K = 0.0112 mol

Step 5: Calculate the moles of MnO₂ needed to produce 0.0112 moles of Cl₂

The molar ratio of MnO₂ to Cl₂ is 1:1. The moles of MnO₂ needed are 1/1 × 0.0112 mol = 0.0112 mol.