Question

According to the following reaction, how much energy is evolved during the reaction of 32.5 g B2H6 with excess Cl2? 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

Answer 1

Answer:

1640.3 kJ of energy is released

Explanation:

Step 1: data given

Mass of B2H6 = 32.5 grams

Cl2 = in excess

The molar mass of B2H6 is 27.67 g/mol.

ΔH°rxn = -1396 kJ

Step 2: The balanced equation

B2H6(g) + 6 Cl2(g) → 2 BCl3(g) + 6 HCl(g)   -1396 kJ

This means when 1 mol of B2H6 reacts with 6 moles of Cl2 there will 1396 kJ of energy be released

Step 3: Calculate the moles of B2H6

Moles B2H6 = mass B2H6 / molar mass B2H6

Moles B2H6 = 32.5 grams / 27.67 g/mol

Moles B2H6 = 1.175 moles

Step 4: Define the limiting reactant

Since Cl2 is in excess, B2H6 will be the limiting reactant.

For 1 mol B2H6 we need 6 moles Cl2 to produce  2 moles BCl3 an 6 moles HCl

For this reaction 1396 kJ of energy will be given off

For 1.175 moles B2H6 the amount of energy released will be:

1.175 * 1396 = 1640.3 kJ

1640.3 kJ of energy is released

NOTE: the negative sign in the question show the energy is released

Answer 2

Answer:

1640 kJ are involved in the reaction

Explanation:

In the reaction:

B₂H₆(g) + 6Cl₂(g) → 2 BCl₃(g) + 6HCl(g)

1 mol of B₂H₆(g) with 6 moles of Cl₂(g) produce 1396 kJ of energy.

Now if 32.5g of B₂H₆(g) react with excess Cl₂(g), moles involved in reaction are:

$32.5g B_2H_6\frac{1mol}{27.67g} = 1.175 moles$

If 1 mol produce 1396kJ of energy, 1.175 moles produce:

$1.175mol\frac{1396kJ}{1mol} = 1640kJ$

Thus, 1640 kJ are involved in the reaction