How much energy is evolved during the rxn of 32.5g of B2H6 and 72.5g of Cl2?

Question

Molar mass of B2H6 is 27.62 g/mol

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

Delta H rxn= -1396 kJ

Answer 1

First, determine how many moles you have of each reactant and what the limiting reagent will be. Here, 32.5 g B₂H₆ divided by 27.62 g/mol is 1.18 moles B₂H₆.

72.5 g Cl₂ divided by 70.9 g/mol is 1.02 moles Cl₂

(look at the periodic table. 2 atoms of Cl will have the mass of 70.9 amu. Replace amu with g/mol and you're set)

One way to find the limiting reagent is to calculate how much of a certain product you can get from the starting amount of each reagent assuming excess (enough) of the others.

For this problem, I would notice that for every mole of B₂H₆ reacted, you need 6 moles of Cl₂. We clearly have less Cl₂ than 6*1,18, so Cl₂ will limit the reaction.

Then you set up a ratio: amount theoretical Cl₂ amount "experimental" Cl₂

----------------------------- = ------------------------------------

theoretical energy "experimental" energy

6 moles 1.02 moles

---------- = ----------

-1396 kJ x

Solve for x (remember the unit is kJ and the sign is negative)

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