chemical calculation...

The relative stomic mass of element X is 74.9. It forms an oxide containing 24.3% of oxygen by mass. What is the mole ratio of X to oxygen in the oxide?

( Relative atomic mass: O=16.0)

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2 個解答

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  • 匿名
    1 十年前
    最愛解答

    Let 1:y be the mole ratio of X to oxygen in the oxide

    [ (16*y) / (16*y + 74.9) ] x 100% = 24.3%

    16y = 3.888 + 18.2007

    y = 1.38 ( corr. to 3sig. fig.)

    1 : 1.38 ~ 2 : 3

    the mole ratio of X to oxygen in the oxide is 2:3

    資料來源: Me ( 可能會計錯)
  • 1 十年前

    let 100 g be the mass of the oxide

    x=100-24.3=75.7g

    no. of moles of x= 75.7/74.9=1.0107

    no. of moles of o2= 24.3/32= 0.7594

    relative no of moles= 4:3

    i think the ratio should be 4:3

    it may be wrong i am not sure

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