Answer:

Correct Answer: A

Solution:

The balanced chemical equation is $ \underset{\begin{smallmatrix} \\ \\ 24g \end{smallmatrix}}{\mathop{Mg}},+\underset{\begin{smallmatrix} \\ 16g \end{smallmatrix}}{\mathop{\frac{1}{2}O _2}},\xrightarrow{,}\underset{\begin{smallmatrix} \\ \\ 40g \end{smallmatrix}}{\mathop{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 $ \frac{16}{24}\times 0.67g,O _2=0.67g,O _2. $ But only 0.56 g $ O _2 $ is available which less than 0.67 g is.

Thus, $ O _2 $ is the limiting reagent. Further, 16 g $ O _2 $ reacts with Mg 24 g.

$ \therefore $ 0.56 g $ O _2 $ will react with Mg $ =\frac{24}{16}\times 0.56=0.84g $

$ \therefore $ Amount of Mg left unreacted = 1.0 - 0.84g Mg = 0.16g Mg Hence, Mg is present in excess and 0.16 g Mg is left behind unreacted.