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Smile 發問時間: 科學及數學化學 · 2 個月前

Chemistry problem: how to do, thanks?

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1 個解答

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  • SC147
    Lv 6
    2 個月前
    最佳解答

    a)

    mass of 1 mole of MnO₂ = 54.9+2(16) = 86.9 g

    mass of 1 mole of HCl = 1+35.5 = 36.5 g

    217 g of MnO₂ contains: 217/86.9=2.497 moles

    274 g of ‬HCl contains: 274/36.5= 7.507 moles

    In this reaction, 

    1 mole of MnO₂ reacts with 4 moles of HCl ,

    y moles of MnO₂ will react with 7.507 moles of HCl

    ∴ y = 7.507/4 = 1.877 (moles) of MnO₂ 

    Since there are 2.497 moles of MnO₂, 

    MnO₂ will be in excess.

    So, HCl is the limiting reactant.

    b) 4 moles of HCl produce 1 mole of Cl₂  

    =>7.507 moles HCl produce 7.507/4=1.877 mole Cl₂  

    mass of 1 mole of Cl₂ = 2(35.5) = 71 g

    ∴mass of Cl₂ produced=71(1.877)=133.3 g(to 1 dec.pl.)‬

    c) no. of moles of MnO₂ used = 1.877 

    ∴no. of moles of MnO₂ in excess=2.497-1.877=0.62

    ∴mass of excess reagent (MnO₂) left after reaction

    = 86.9 (0.62) = 53.88 g (to 2 dec.pl.)

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