Question

What is the percentage yield if `729*g` of tetrabromoethane is prepared from `60.3*g` acetylene and stoichiometric bromine?

Answer 1

We assess the stoichiometric reaction:

`HC-=CH(g) + 2Br_2(l) rarrBr_2C-CBr_2H`

Explanation:

And thus there is 1:1 stoichiometry between the acetylene reactant and the tetrabromo product:

Now `"%yield"="Moles of product"/"Moles of reactant"xx100%`

And here, `"%yield"` `=` `((729.0*g)/(345.65*g*mol^-1))/((60.3*g)/(26.04*g*mol^-1))xx100%=91.1%`

The calculation for `"yield"` is simplified a bit here in that there were excess bromine, and 1:1 stoichiometry between the organic reactant, and the organic product. Do you see what is going on?

I would not be very happy if I had to do a reaction with almost a half kilo of elemental bromine. It is one of the most corrosive substances to handle in a lab, and can cause horrific burns.