A) zero
B) - 2.3 kJ
C) 2.2kJ
D) 0.22kJ
Correct Answer: B
Solution :
\[\underset{55.85\,\,g}{\mathop{Fe(s)}}\,+2HCl(aq)\xrightarrow{{}}FeC{{l}_{2}}(aq)+\underset{1\,\,mol}{\mathop{{{H}_{2}}(g)}}\,\] \[n=\frac{50}{55.85}=0.895\,\,mol\] \[W=-nRT-0.895\times 8.314\times {{10}^{-3}}\times 298\] \[=-2217.42\times {{10}^{-3}}=-2.2\,\,kJ\] The reaction mixture does - 2.2 kJ of work driving back the atmosphere.You need to login to perform this action.
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