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1.0 g of Mg is burnt with 0.28 g of O(2)...

1.0 g of Mg is burnt with 0.28 g of `O_(2)` in a closed vessel. Which reactant is left in excess and how much ?

A

Mg , 5.8 g

B

Mg , 0.58 g

C

`O_2` , 0.24 g

D

`O_2` , 2.4 g

Text Solution

Verified by Experts

The correct Answer is:
B
Balanced reaction can be given as
`2Mg +O_2rarr2MgO`
32 g of `O_2` is required to burn 48 g Mg
So, 0.28 g `O_2` will be required for `48/32 xx0.28` g Mg
=0.42 g of Mg
Thus, Mg will remain in excess = 1 - 0.42 = 0.58 g
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Knowledge Check

  • 1 g of Mg is burnt with 0.28 g of O_2 in a closed vessel which reactant is left in excess and how much?

    A
    `Mg, 5cdot8g`
    B
    `Mg, 0cdot58g`
    C
    `O_2,0cdot24g`
    D
    `O_2,2cdot4g`

  • 1.0 g of magnesium is burnt with 0.56 g O_2 in a closed vessel. Which reactant is left in excess and how much? (At. Wt. Mg=24,O=16)

    A
    Mg, 0.16 g
    B
    `O_2, 016g`
    C
    Mg, 0.44g
    D
    `O_2, 0.28g`

  • When 0.5 g of sulphur is burnt to SO_(2):4.6 kJ of heat is liberated. What is the enthalpy of formation of sulphur dioxide ?

    A
    `-147.2 kJ`
    B
    `+147.2 kJ`
    C
    `+294.4 kJ`
    D
    `-294.4 kJ`
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