1 g of Mg is burnt in a closed vessel containing 0.5 g of `O_(2)`. Which of the following statements is correct ?

Options

A. `O_(2)` is the limiting reagent and Mg is in excess by 0.25 g
B. Mg is the limiting reagent and is in excess by 0.5 g
C. `O_(2)` is the limiting reagent and is in excess by 0.25 g
D. `O_(2)` is the limiting reagent and Mg is in excess by 0.75 g

Answer

A

Explanation

`underset(2xx24)(2Mg)+underset(2xx16)(O_(2))rarr2MgO`
48 g of Mg require `O_(2)=32` g
1g of Mg require `O_(2)=(32)/(48) = 0.66 g`
`O_(2)` actually available = 0.5 g
This means that `O_(2)` is the limiting reactant
0.5 g of `O_(2)` will react with `Mg = (48)/(32) xx 0,5 = 0.75 g`
Excess of `Mg = 1 -0.75 =0.25 g`.