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March 27, 2017

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Table salt, NaCl, and sugar C12H22O11 are accidentally mixed. A 5.50g sample is burned and 3.60g of CO2 is produced. What was the mass percentage of the table salt mixture?

  • Chemistry - ,

    MC12H22O11=342
    MCO2=44
    C12H22O11+xO2=12CO2+...
    Ratio of mass of C12H22O11 to CO2
    =342:12*44
    =342:484
    Mass of C12H22O11 that produces 3.6g CO2
    =3.6*(342/484)
    =2.544g
    Mass ratio NaCl:C12H22O11
    =(5.5-2.544):2.544
    =2.956:2.544
    Calculate mass percentage and reduce to 3 significant figures.

  • oops--Chemistry - ,

    3.60 x (342/12*44) = 3.6 x (342/528) = 2.33 g C12H22O11 to produce 3.60 g CO2.

    Then 5.50-2.33 = 3.17 g NaCl and
    %NaCl = (3.17/5.50)*100 = ?

  • Chemistry - ,

    57.6 %

  • Chemistry - ,

    3.60 x (342/12*44) = 3.6 x (342/528) = 2.33 g C12H22O11 to produce 3.60 g CO2.

    Then 5.50-2.33 = 3.17 g NaCl and
    %NaCl = (3.12/5.59)*100 = ?

  • Chemistry - ,

    70.44

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