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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