Question

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 ?
(At. wt. Mg = 24 ; O = 16)

• A. Mg, 0.16 g
• B. $O_2$, 0.16 g
• C. Mg, 0.44 g
• D. $O_2$, 0.28 g

Answer 1

Answer: (A)

The balanced chemical equation is $\underset{24g}{Mg} + \underset{16g}{\frac{1}{2}O_2} \longrightarrow \underset{40 g}{MgO}$
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.