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:
[tex]32.5g B_2H_6\frac{1mol}{27.67g} = 1.175 moles[/tex]
If 1 mol produce 1396kJ of energy, 1.175 moles produce:
[tex]1.175mol\frac{1396kJ}{1mol} = 1640kJ\\[/tex]
Thus, 1640 kJ are involved in the reaction
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
