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

Answer:

The enthalpy of formation of CaF₂ is -1224.4 kJ.

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:

$n_{CaF_{2}} = \frac{m}{M}$

Where:

m: is the mass of CaF₂ = 16 g

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

$n_{CaF_{2}} = \frac{m}{M} = \frac{16 g}{78.07 g/mol} = 0.205 moles$

Now, the enthalpy of formation of CaF₂ is:

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

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

I hope it helps you!