A) 0.333
B) 0.011
C) 0.029
D) 0.044
Correct Answer: C
Solution :
\[PbO+2HCl\xrightarrow{{}}PbC{{l}_{2}}+{{H}_{2}}O\] \[{{n}_{PbO}}=\frac{6.5}{223}=0.029\] \[{{n}_{HCl}}=\frac{3.2}{36.5}=0.087\]| PbO gives \[PbC{{l}_{2}}\] | HCl gives \[PbC{{l}_{2}}\] |
| Moles 0.029 0.029 | 0.087 0.048 |
| \[\Rightarrow \]\[{{n}_{PbC{{l}_{2}}}}=0.029\] |
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