Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid,...

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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid,...

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 175 mL Cl2(g) at 25 °C and 785 Torr ?

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Answer 1

Answer #1

PV = nRT

where, P = pressure = 785 torr = 1.03 atm

V = volume = 175 mL = 0.175 L

n = number of moles of Cl2

R = Gas constant

T = temperature = 25 + 273 = 298 K

1.03 * 0.175 = n * 0.0821 * 298

0.180 = n * 24.5

n = 0.180 / 24.5 = 0.00735 mol

Therefore, the number of moles of Cl2 = 0.00735 mol

From the balanced equation we can say that

1 mole of Cl2 produced by 1 mole of MnO2 so

0.00735 mole of Cl2 will be produced by

= 0.00735 mole of Cl2 *(1 mole of MnO2 / 1 mole of Cl2)

= 0.00735 mole of MnO2

mass of 1 mole of MnO2 = 86.9368 g

so the mass of 0.00735 mole of MnO2 = 0.639 g

Therefore, the mass of MnO2 required would be 0.639 g

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Answer 2

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