CHEMISTRY| MOLE CONCEPT | Question 09


90.0 of FeCL3 react with 52.0g of H2
i. What is the limiting reactant?
ii. What is the mass of HCL produced?
iii. What mass of excess reactant remain after the reaction?


  • DATA GIVEN


    Mass of FeCl3 = 90.0g
    Mass of H2S = 52.0g
    Asked:
    i. The limiting reactant
    ii. Mass of HCl produced
    iii. Mass of excess reactant remain after reaction
    Equation for the reaction
    2FeCl3 + 3H2S → F2S3 + 6HCl

  • i/ The limiting reactant


    To convert mass into mole
    Mole for FeCl3 = mass /Molar mass
    = 90.0g /162.5
    = 0.5538 mole
    Mole ratio = mole /Mole coefficient
    = 0.5538 mole/2
    = 0.2769
    Mole of H2S = mass /Molar mass
    = 1.529 mole /3 mole
    = 0.509

    The limiting reactant is FeCl3
  • ii/ To find the excess mass of HCL produced


    2 moles of FeCl3 Ξ 6 moles of HCl
    0.5538 mole of FeCl3 Ξ ?? of HCl
    = (0.5538 × 6) moles2 /2 mole
    = 1.6614 mole of HCl
    = mass of HCl produced
    Mass = mole × molar mass
    = 1.6614 mole × 36.5g/mole
    = 60.6411g
    60.6411g of HCl was produced.
  • iii/ calculate the mass of H2S remain


    Mass reacted from mole equivalent
    2moles of FeCl3 Ξ 3 moles of H2S
    0.5538 moles of FeCl3 Ξ ??
    = 0.5538 mole × 3 moles /2 moles
    = 0.8307 mole of H2S
    To convert 0.8307 moles into mass
    Mass = mole × molar mass
    = 0.8307 mole × 34g/mole
    = 28.2438g
    Mass remain of H2S = mass of origin – mass reacted
    = 52.0g – 28.2438g
    = 23.7562g of H2S remain unreacted

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