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,0.44g`
D. `O_(2),0.28g`

Answer

A

Explanation

`underser(" = 48")underset(2xx24g)(2Mg)+underset(32g)(O_(2))rarr2MgO`
1 g of Mg needs `O_(2)=(32)/(48)g=(2)/(3)g=0.67g` which is not present.
`0.56` g of `O_(2)` will react with Mg
`=(48)/(32)xx0.56=0.84`
Thus, Mg is left in excess and amount left
`=1-0.84g=0.16g`