Question

A 8.22-g sample of solid calcium reacted in excess fluorine gas to give a 16-g sample of pure solid CaF2. The heat given off in this reaction was 251 kJ at constant pressure. Given this information, what is the enthalpy of formation of CaF2(s)

Answer 1

The enthalpy of formation of CaF₂ is -1224.4 kJ.

Step-by-step explanation:

The enthalpy of formation of CaF₂ can be calculated as follows:

`\Delta H_(f) = \frac{q}{n_{CaF_(2)}}`

Where:

q: is the heat liberated in the reaction = -251 kJ

The number of moles of CaF₂ is:

`n_{CaF_(2)} = (m)/(M)`

Where:

m: is the mass of CaF₂ = 16 g

M: is the molar mass of CaF₂ = 78.07 g/mol

`n_{CaF_(2)} = (m)/(M) = (16 g)/(78.07 g/mol) = 0.205 moles`

Now, the enthalpy of formation of CaF₂ is:

`\Delta H_(f) = \frac{q}{n_{CaF_(2)}} = (-251 \cdot 10^(3) J)/(0.205 moles) = -1224.4 kJ/mol`

Therefore, the enthalpy of formation of CaF₂ is -1224.4 kJ.

I hope it helps you!