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When 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of \({\bf{FeC}}{{\bf{l}}_{\bf{2}}}\)(s) and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of \({\bf{FeC}}{{\bf{l}}_{\bf{2}}}\)(s) is produced?

Short Answer

Expert verified

Enthalpy change = - 344kJ for 1 mol of \({\rm{FeC}}{{\rm{l}}_{\rm{2}}}\).

Step by step solution

01

Number of moles

The reaction is:

\({\rm{2Fe}}\left( {\rm{s}} \right){\rm{ + 2C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{FeC}}{{\rm{l}}_{\rm{2}}}\left( {\rm{s}} \right)\)

The number of moles is evaluated as:

\({\bf{Number of moles = }}\frac{{{\bf{Given mass}}}}{{{\bf{Molar mass}}}}\)

We have

\(\frac{{{\rm{1}}{\rm{.42g}}}}{{{\rm{55}}{\rm{.8g/mol}}}}{\rm{ \times 2 = 0}}{\rm{.04 mol of iron is available}}\)

\(\frac{{{\rm{3}}{\rm{.22g}}}}{{{\rm{126}}{\rm{.75g/mol}}}}{\rm{ \times 2 = 0}}{\rm{.05 mol of FeC}}{{\rm{l}}_{\rm{2}}}{\rm{ is available}}\)

02

Calculation of enthalpy change

The reaction uses 2 moles of \({\rm{FeC}}{{\rm{l}}_{\rm{2}}}\)and the conversion factor is - 8.60 / 0.05 mol of \({\rm{FeC}}{{\rm{l}}_{\rm{2}}}\). So we have,

\(\begin{array}{l}{\rm{ = }}\frac{{{\rm{ - 2 moles of FeC}}{{\rm{l}}_{\rm{2}}}{\rm{\;}} \times {\rm{8}}{\rm{.60}}}}{{{\rm{0}}{\rm{.05 FeC}}{{\rm{l}}_{\rm{2}}}{\rm{\;}}}}\\{\rm{ = - 344 kJ}}\end{array}\)

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