Answer:
2Li(s)+Cl2(g)→2LiCl(s)
C(s,graphite)+ O2(g)→ CO2(g)
6Li(s)+12Cl2(g)→6LiCl(s)
Explanation:
Hello,
The following reactions:
2Li(s)+Cl2(g)→2LiCl(s)
C(s,graphite)+ O2(g)→ CO2(g)
6Li(s)+12Cl2(g)→6LiCl(s)
Are characterized by the presence of just elements in the reagents, thus, their enthalpy of formation is cero, so the change in the enthalpy of reaction is equal to the enthalpy of formation of the products.
Best regards.