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27 gm of Al reacts with excess of oxygen...

27 gm of `Al` reacts with excess of oxygen to give 4.59 gm of `Al_(2)O_(3)`. Calculate percentage yield of reaction.

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The number of oxygen atoms present in 20.4 g of Al_(2)O_(3) are equal to the number of :

A sample of 10gm of H_(2) reacts with sufficient amount of oxygen to form 106gm of H_(2)O(l) and H_(2)O_(2)(l) . Calculate mass % of H_(2)O .

Knowledge Check

  • Al_(2)O_(3) reacts with

    A
    only water
    B
    only acids
    C
    only alkalis
    D
    both acids and alkalis

  • 10 g of S reacts with excess of O_2 to form 15 g of SO_2. The % yield of the reaction is

    A
    0.25
    B
    0.5
    C
    0.75
    D
    1

  • If half mole of oxygen combine with Al to form Al_(2)O_(3) the weight of Al used in the reaction is:

    A
    `27 g`
    B
    `40.5 g`
    C
    `54 g`
    D
    `18 g`

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