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Stoichiometry

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How many grams of oxygen are required to completely burn 85.6 grams of methane (CH4)?

CH4 + O2 → H2O + CO2

  • Stoichiometry -

    85.6g CH4 = 5.34 moles
    each mole CH4 reacts with 1 mole of O2
    so, you need 5.34 moles O2 = 85.39g O2

  • Stoichiometry -

    oops: O2 = 32, not 16, so that's 170.78g

  • Stoichiometry -

    How many grams of 02 are needed to form 17 grams of H2O

  • Stoichiometry -

    n=M/RFM
    n=number of moles RFM= relative formula mass

    n=17/18=0.944
    0.944 moles
    then
    0.944=x/16
    x=0.944*16
    x=15.104
    15.105 grams

  • Stoichiometry -

    dcfd47ur5b

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