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?

Options

A. `Mg,0.16 g`
B. `O_(2),0.16g`
C. `Mg,0.44 g`
D. `O_(2),0.28 g`

Answer

A

Explanation

The balance chemical equation is `{:(Mg,+,(1)/(2)O_(2),to,MgO),(24g,,16g,,40g):}`
From the above equation , it is clear that, 24 g of Mg reacts with 16 g of `O_(2)`.
Thus , 1.0 g of Mg reacts with
`(16)/(24)g` of `O_(2)=0.67g of O_(2)`.
But only 0.56 g of ` O_(2)` is available which is less then 0.67 g . thus , `O_(2)` is the limiting reagent. Further , 16 g of `O_(2)` reacts with 24 g of Mg.
`:.`0.56 g of `O_(2)` will with react with Mg `=(24)/(16)xx0.56=0.84g`
`:.` Amount of Mg left unreacted
=(1.0-0.84) g Mg =0. 16 g Mg