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

Answer

A

Explanation

The balanced chemical equaiton is
`underset (24 " g")(Mg)+underset(16" g")(1/2O_(2))rarrunderset(40g) (MgO)`
From the above equation, it is clear that 24 g Mg reacts
with 16 g `O_(2)`
Thus, `1.0` g Mg reacts with `16/24 O_(2) = 0.67 g O_(2)`
But only `0.56 g O_(2)` is available which is less then `0.67` g.
Thus, `O_(2)` is the limiting reagent.
Further, 16 g `O_(2)` reacts with 24 g Mg.
`therefore 0.56 g O_(2)` will react with Mg `+ 24/16xx0.56= 0.84` g
`therefore` Amount of Mg left unreacted `= 1.0 - 0.84 = 0.16` g Mg
Hence, Mg is present is excess and `0,16` g Mg is left
behind unreacted.