Question

what is the % yield in this experiement

Answer 1

There are no picture, image or data to do this. However, let me put you an example of how you can calculate the %yield in general for a reaction experiment.

Say you have the following reaction:

Cu(NO₃)₂ + PbCl₂ -------------> CuCl₂ + Pb(NO₃)₂

And you want to know how much of Pb(NO₃)₂ is produced (theorycally speaking) from a sample of 27.85 g of Cu(NO₃)₂ in enough PbCl₂ which is aqueous.

First, you need to balance the reaction, and see the relation between the reactant and product. In this case, is already balance so the relation is 1:1, which means that the mole of Cu(NO₃)₂ = moles of Pb(NO₃)₂

calculate the moles of Cu(NO₃)₂ with the molecular weight reported of this compound:

moles = 27.85 g / 187.56 g/mol = 0.1485 moles

this moles are the same moles produced of Pb(NO₃)₂, so the mass (theorical) produced will be:

m = 0.1485 * 331.2 = 49.18 g

This is the theorical mass. but let's suppose that the experimental mass obtained was 37.65 g and you want to know the %yield. Now that we know the theorical mass produced, we can calculate the %yield

%yield = (exp mass / theor mass) * 100

% = (37.65 / 49.18) * 100 = 76.56%

So, in order to calculate the %yield you need to know the theorical amount obtained (in mass, moles or even volume) and the experimental obtained, and then apply the above expression.

Hope this information gives you an idea. If you still have doubts, please provide the actual experiment.