A) 75
B) 96
C) 60
D) 84
Correct Answer: D
Solution :
| Key concept In the given problem we have provided practical yield of \[MgO\]. For calculation of percentage yield of \[MgO,\] we need theoretical yield of \[MgO\]. For this we shall use mole concept. |
| \[MgC{{O}_{3}}(s)\xrightarrow{\,}MgO(s)+C{{O}_{2}}(g)\,\] (i) |
| Moles of \[MgC{{O}_{3}}\] |
| \[\text{=}\frac{\text{weight m gram}}{\,\text{Molecular weight}}\] |
| \[=\frac{20}{84}=0.238mol\] |
| From Eq. (i) |
| 1 mole of \[MgC{{O}_{3}}\] gives \[=\,\,1\,\,mol\,\,MgO\] |
| \[\therefore \] 0.238 mole \[MgC{{O}_{3}}\] will give = 0.238 mol \[MgO\] |
| \[=0.238\times 40\,g\,=9.52g\,MgO\] |
| Now, practical yield of \[MgO=8\,g\] |
| \[\therefore \] \[%\text{purity=}\frac{8}{9.52}\therefore 100=84%\] |
| Alternate, |
| \[\underset{84g}{\mathop{MgC{{O}_{3}}}}\,\xrightarrow{\,}\underset{40g}{\mathop{MgO}}\,+C{{O}_{2}}\] |
| \[\therefore \] \[8\,\,g\,MgO\] will be form from \[\frac{84}{5}g\] |
| \[\therefore \] % purity \[=\frac{84}{2}\times \frac{100}{20}=84%\] |
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