A) \[Mg,0.16g\]
B) \[{{O}_{2}}.0.16g\]
C) \[Mg,0.44g\]
D) \[{{O}_{2}},0.28g\]
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.
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