Answered

Connect with knowledgeable experts and enthusiasts on IDNLearn.com. Join our community to receive prompt and reliable responses to your questions from knowledgeable professionals.

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

Sagot :

Jbferradoidn

Answer:

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

We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com has the solutions you’re looking for. Thanks for visiting, and see you next time for more reliable information.