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

Question

Natalie Anton · Accepted Answer

We assess the stoichiometric reaction:

#HC-=CH(g) + 2Br_2(l) rarrBr_2C-CBr_2H# Consequently, the acetylene reactant and the tetrabromo product have a 1:1 stoichiometry: 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? Having to perform a reaction with nearly half a kilogram of elemental bromine would not be fun. It's one of the most caustic chemicals to work with in a lab and can result in horrifying burns.

Mateo Aguilar · Answer

undefined To calculate the percentage yield, we first need to determine the theoretical yield of tetrabromoethane using stoichiometry, based on the balanced chemical equation for the reaction between acetylene and bromine to form tetrabromoethane.

Once we have the theoretical yield, we can then calculate the percentage yield using the following formula:

Percentage Yield = (Actual Yield / Theoretical Yield) * 100%

Actual yield refers to the amount of product obtained experimentally, while theoretical yield is the maximum amount of product that could be obtained under ideal conditions.

Given that the reaction occurs with stoichiometric amounts of bromine, we'll need to calculate the amount of tetrabromoethane that can be produced from the limiting reactant, which is acetylene.

We'll need the balanced chemical equation for the reaction between acetylene (C2H2) and bromine (Br2) to form tetrabromoethane (C2H2Br4). Once we have that, we can calculate the theoretical yield of tetrabromoethane.