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1g of Mg is burnt with 0.28g of O(2) in ...

1g of Mg is burnt with 0.28g of `O_(2)` in a closed vessel. Which reactant is left in excess and how much?

A

Mg,5.8g

B

Mg,0.58g

C

`O_(2),0.24g`

D

`O_(2),2.4g`

Text Solution

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The correct Answer is:
B
`2Mg(s)+O_(2)(g)rarr2MgO`
48g 32g 80g
32g `O_(2)-=48gMg`
`0.28g O_(2)-=(48)/(32)xx0.28g Mg`
=0.42g Mg
Excess `Mg=1-0.42=0.58g`
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