Suppose 33.5 g of ClF3(g) and 29.1 g of Li(s) are mixed and allowed to react at atmospheric pressure and 25 C until one of the reactants is used up, producing LiCl(s) and LiF(s). Using the NIST value for the enthalpy of formation of ClF3 (webbook.nist.gov), calculate the enthalpy change for this reaction, keeping in mind that this is a negative number for an exothermic process.

To calculate the enthalpy change for the reaction, we need to use the balanced chemical equation for the reaction, the enthalpy of formation values for the products and reactants, and apply the principle of Hess's Law.

1. Write the balanced chemical equation for the reaction:
2ClF3(g) + 6Li(s) → 6LiF(s) + Cl2(g)

2. Determine the molar masses of ClF3 and Li:
Molar mass of ClF3 = 33.5 g/mol
Molar mass of Li = 29.1 g/mol

3. Convert the given masses of ClF3 and Li to moles:
Moles of ClF3 = mass / molar mass = 33.5 g / 88.45 g/mol = 0.3787 mol
Moles of Li = mass / molar mass = 29.1 g / 6.941 g/mol = 4.1858 mol

4. Determine the limiting reactant:
Since the stoichiometric ratio between ClF3 and Li is 2:6 (1:3), we can see that ClF3 is the limiting reactant since we have less moles of it. This means that it will be completely consumed in the reaction, and the reaction will stop once all the ClF3 is used up.

5. Calculate the moles of the products formed:
From the balanced equation, 2 moles of ClF3 react to form 6 moles of LiF:
Moles of LiF = (moles of ClF3) * (6 moles of LiF / 2 moles of ClF3) = 0.3787 mol * (6/2) = 1.1361 mol
Moles of Cl2 = 0.3787 mol * (1 mole of Cl2 / 2 moles of ClF3) = 0.1894 mol

6. Look up the enthalpy of formation values for LiCl and LiF:
Using the NIST Chemistry WebBook (webbook.nist.gov), we find:
Enthalpy of formation of LiCl(s) = -408.28 kJ/mol
Enthalpy of formation of LiF(s) = -618.08 kJ/mol

7. Apply Hess's Law to calculate the enthalpy change for the reaction:
Enthalpy change = (Σ n * ∆Hf(products)) - (Σ n * ∆Hf(reactants))
= (6 mol * (-618.08 kJ/mol)) + (1 mol * (-408.28 kJ/mol)) - (0 mol * ∆Hf(ClF3)) - (4.1858 mol * 0 kJ/mol)
= -3708.48 kJ

8. The enthalpy change for this reaction is -3708.48 kJ, which indicates an exothermic process.

Note: Hess's Law allows us to calculate the enthalpy change for a reaction by subtracting the sum of enthalpies of formation of the reactants from the sum of enthalpies of formation of the products. The enthalpy of formation values for the substances involved in the reaction are obtained from reliable sources such as the NIST Chemistry WebBook.