1.

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

Answer»

Mg,5.8g
Mg,0.58g
`O_(2),0.24g`
`O_(2),2.4g`

SOLUTION :`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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