How much energy is released when 67.04g of phosphorous is reacted with 10.20g of chlorine? ___ P + ___ Cl2 ___ PCl3 ΔH = -574 kJ

26.78 kJ

Explanation:

To solve the problem, we have to first write the stoichiometric coefficients in the chemical equation:

2P + 3Cl₂ → 2PCl₃

With these coefficients, we have the same number of atoms of each chemical element on both sides: 2 atoms of P, 6 atoms of Cl.

According to the equation, 2 moles of phosphorous (P) react with 3 moles of chlorine (Cl₂), and 574 kJ of energy are released. We have to figure out which is the limiting reactant. For this, we convert the mass into moles by using the molar mass(MM):

MM(P) = 30.9 g/mol

67.04 g P/(30.9 g/mol) = 2.17 mol P

MM(Cl₂) = 35.4 g/mol x 2 = 70.8 g/mol

10.20 g Cl₂/(70.8 g/mol) = 0.14 mol Cl₂

Now, we multiply the actual moles of P (the amount we have for the reaction) by the stoichiometric ratio given by the chemical equation (3 mol Cl₂/2 mol P):

2.17 mol P x (3 mol Cl₂/2 mol P) = 3.25 mol Cl₂

To completely react 67.04 g P, we need 3.25 mol of Cl₂, and we have only 0.14 moles of Cl₂, so the limiting reactant is Cl₂.

Now, we use the limiting reactant to calculate the energy released from the reaction. The energy released per mole of Cl₂ is:

ΔH/(3 mol Cl₂) = -574 kJ/3 mol Cl₂= 191.3 kJ/mol Cl₂

Finally, we multiply the energy released per mole of Cl₂ by the number of moles of Cl₂ we have:

0.14 mol Cl₂ x 191.3 kJ/mol Cl₂ = 26.78 kJ