PbS + 2PbO → 3Pb + SO2 a. Balance the equation. b. How many moles of Pb will form from 17 moles of PbO (excess PbS)? 57.33g PbO x 1 mole of

Question

PbS + 2PbO → 3Pb + SO2 a. Balance the equation. b. How many moles of Pb will form from 17 moles of PbO (excess PbS)? 57.33g PbO x 1 mole of PbO/ c. How many moles of PbO are needed to make 27.3 moles of SO2 (excess PbS)? d. Give 17.6 moles of PbS and 36 moles of PbO, which is the limiting reactant?

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Dulcie 4 years 2021-08-22T06:03:48+00:00 1 Answers 26 views 0

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  1. Gerda
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    2021-08-22T06:04:57+00:00
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    Answer:

    Explanation:

    a )

    PbS      +     2PbO   →     3Pb    +     SO₂

                      2 mole       3 moles       1 moles

    b )

    2 moles of PbO is required to produce 3 moles of Pb

    17 moles of PbO is required to produce 3/2 x 17 moles of Pb

    = 3/2 x 17 moles of Pb

    = 25.5 moles of Pb .

    c )

    1 mole of  SO₂ needs 2 moles of PbO

    27.3 mole of  SO₂ needs 2 x 27.3  moles of PbO

    PbO needed = 2 x 27.3 moles

    = 54.6 moles .

    d )

    PbS      +     2PbO   →     3Pb    +     SO₂

    1 mole       2 mole    

    1 mole  of  PbS requires 2 moles of PbO

    17.6 mole  of  PbS requires 2 x 17.6  moles of PbO

    PbO required = 2 x 17.6 moles

    = 35.2 moles .

    PbO available = 36 moles . So PbO is in excess .

    Hence PbS is the limiting reagent .

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