1.

1.0 g of magnesium is burnt with 0.56 g O_(2) in a closed vessel. Which reaction is left in excess and how much ? ( At. wt. Mg = 24, O = 16)

Answer»

`MG, 0.16 g `
`O_(2), 0.16 g`
`Mg, 0.16 g`
`O_(2), 0.28 g`

Solution :`M g + 1/2 O_(2) RARR MgO`
`(1.0)/(24) (0.56)/(32)`
Oxygen is LIMITING reagent so `(0.5)/(12) - x(0.07)/(4) - x/2`
`(0.07)/(4) - x/2 = 0`
`x=(0.07)/(2)`
excess `Mg=(0.5)/(12) - (0.07)/(2)` mole
`rArr 1-0.7xx12= 0.16g`


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