According to the following reaction, how much energy is evolved during the reaction of 0.111 mol B2H6 and 0.252 mol Cl2 (Both gases are initially at STP)?

Question

According to the following reaction, how much energy is evolved during the reaction of 0.111 mol B2H6 and 0.252 mol Cl2 (Both gases are initially at STP)? The molar mass of B2H6 is 27.67 g/mol.

            B2H6(g) + 6 Cl2(g) → 2 BCl3(g) + 6 HCl(g)      ΔH°rxn = -1396 kJ

	A.	58.7 kJ
	B.	156 kJ
	C.	215 kJ
	D.	352 kJ

Answer 1

Problem:

Find out how much energy is produced with 0.111 mol B2H6 and 0.252 mol Cl2.

Steps:

1) Find the limiting reactant using the chemical equation above

2) Use the limiting reactant to calculate how much energy is evolved.

Step 1:

To calculate limiting reactant, pick either B2H6 or Cl2, and find out if you're short of the other reactant. For this example, I'll pick Cl2.

0.252 mol Cl2 × ( 1 mol B2H6 / 6 mol Cl2) = 0.042 mol B2H6 required.

The problem statement says that we have 0.111 mol B2H6, which is far greater than 0.042 mol B2H6 needed for the reaction. Therefore, Cl2 is the limiting reagent.

Step 2:

Now that we know Cl2 is the limiting reagent, we can calculate how much energy is evolved using the same chemical equation:

0.252 mol Cl2 × (1396 kJ / 6 mol Cl2) = 58.7 kJ, or answer A.

Answer 2

B2H6(g) + 6 Cl2(g) → 2 BCl3(g) + 6 HCl(g)      ΔH°rxn = -1396 kJ

Find the limiting reactant:

B2H6 = 0.111 mol

Cl2 = 0.252 mol/6 = 0.042 so Cl2 is limiting

0.252 mol Cl2 x 1 mol B2H6/6 mol Cl2 x 1396 kJ/mol B2H6 = 58.6 kJ <-- ANSWER A