| For the reaction, |
| \[CaO+2HCl\xrightarrow{{}}CaC{{l}_{2}}+{{H}_{2}}O\] |
| \[1.23\,g\]of \[CaO\]is reacted with excess of hydrochloric acid and \[1.85\text{ }g\] of \[CaC{{l}_{2}}\] is formed. What is the per cent yield? |
A) 76.1
B) 86.3
C) 95.1
D) None of these
Correct Answer: A
Solution :
| The balanced equation is |
| \[\underset{\begin{smallmatrix} 1mol \\ 56g \end{smallmatrix}}{\mathop{CaO}}\,+2HCL\to \underset{\begin{smallmatrix} 1mol \\ 111g \end{smallmatrix}}{\mathop{CaC{{l}_{2}}}}\,+{{H}_{2}}O\] |
| \[56g\,of\,CaO\,produces\,CaC{{l}_{2}}=111g\] |
| \[1.23g\,\,of\,CaO\,\]will produce\[CaC{{l}_{2}}=\frac{111}{56}\times 1.23\] |
| Thus, theoretical yield = 2.43 g |
| Actual yield = 1.85 g |
| Per cent yield \[=\frac{1.85}{2.43}\times 100=76.1\] |
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