Can someone explain the steps to this? 10pts to Best!?

This is the question:

Chlorine gas can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid:

MnO2(S) + 4HCl(aq) => MnCl2(aq) + 2H2O(l) + Cl2(g) How much MnO2 should be added to excess HCl to obtain 275 ml of chlorine gas at 5.0 oC and 650 mm Hg?

I know the answer is 0.896g but what are the steps to getting to this??

1 Answer

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  • 8 years ago
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    determine moles of Cl2-

    PV=nRT. where n= no. of moles, T= temp. oK

    n= PV/RT = 0.855 atm x 0.275 L/0.082 x 278.15 oK

    n= 0.01

    from bal. rxn., 1 mole of Cl2 is formed from 1 mole of MnO2, therefore 0.01 mole MnO2 required

    0.01 moles MnO x 86.94 g/mole MnO2= 0.869 g

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