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How much MnO2(s) should be added to excess HCl(aq) to obtain 295 mL of Cl2(g) at...

How much MnO2(s) should be added to excess HCl(aq) to obtain 295 mL of Cl2(g) at 25 °C and 765 Torr?

(MnO2 and Cl2 have a one to one ratio)

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Answer #1

MnO2   + 4 HC l -----------------------> MnCl2 + 2 H2O + Cl2

volume = 295 mL = 0.295 L

temperature = T = 25 + 273 = 298 K

pressure = P = 765 torr = 1.007 atm

moles of Cl2 = P V / R T

                       = 1.007 x 0.295 / 0.0821 x 298

                       = 0.0121

moles of Cl2 = moles of MnO2

moles of MnO2 = 0.0121

molar mass of MnO2 = 86.93 g / mol

mass of MnO2 = 0.0121 x 86.93

                        = 1.06 g

mass of MnO2 needed = 1.06 g

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