Chlorine can be prepared in the laboratory by the reaction of manganese dioxide

Question

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) +2H20(l)+ Cl(g)

How much MnO2(s) should be added to excess HCL(aq) to obtain 285mL of Cl2(g) at 25C and 705 Torr?

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) +4HCI(aq) MnCl2(aq) +2H2O(I) + Cl2 How much MnO2(s) should be added to excess HCI(aq) to obtain 285 mL of Cl2(g) at 25 °C and 705 Torr? Number g Mno,

Explanation / Answer

The balanced equation is

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

thus one mole of MnO2 produces one mole of Cl2 gas.

we need to get 285mL of chlorine gas at 25C and 705 torr.

P of CL2 = 705torr = 705torr x 1atm/ 760torr

=0.9276 atm

T of Cl2 = 25C = 25+273

= 298K

V of Cl2 = 285mL = 285mL x1L /1000mL

= 0.285L

Using the ideal gas equation Pv = nRt

We calculate the moles of Cl2 required

moles n = PV/RT

= 0.9276 atm x 0.285L /0.0821L.atm/mol.K x 298K

=0.01080 mol

Now from the stoichiometric equation

1mole of MnO2 produces one mole of Cl2

thus 0.01080molof cl2 is produced from 0.01080 mol of MnO2.

Mass of MnO2 = moles x mollar mass

= 0.01080 molx 87g/mol

= 0.94g

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