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

May 18, 2017

We assess the stoichiometric reaction:

$H C \equiv C H \left(g\right) + 2 B {r}_{2} \left(l\right) \rightarrow B {r}_{2} C - C B {r}_{2} H$

#### 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, $\text{%yield}$ $=$ ((729.0*g)/(345.65*g*mol^-1))/((60.3*g)/(26.04*g*mol^-1))xx100%=91.1%

The calculation for $\text{yield}$ 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.