A) 0.044
B) 0.333
C) 0.011
D) 0.029
Correct Answer: D
Solution :
Key Idea: The reagent which is present in smaller quantity is called the limiting reagent and the moles of product depends on it and \[\text{number}\,\text{of}\,\text{moles}\,\text{=}\frac{\text{weight}}{\text{molecular}\,\text{weight}}\] \[\begin{matrix} PbO & + & 2HCl\xrightarrow[{}]{{}} & PbC{{l}_{2}} & + & {{H}_{2}}O \\ 207.2+16 & {} & 2(35.5+1) & 207.2+71 & {} & {} \\ =223.2 & {} & =73 & =278.2 & {} & {} \\ \end{matrix}\] Here, 1 mole of PbO reacts with 2 moles of HCl, thus PbO is the limiting reagent \[\because \] 223.2 g PbO gives \[PbC{{l}_{2}}=278.2g\] \[\therefore \]6.5 g PbO will give \[PbC{{l}_{2}}\] \[=\frac{278.2}{223.2}\times 6.5g\] \[=\frac{278.2\times 6.5}{223.2\times 278.2}mol\] \[=0.029\,mol\]
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