find percent yield

Question

What is the percent yield of a reaction between 25.0 mL of 0.541 M silver nitrate and 10.0 mL of 3.00 M hydrobromic acid (HBr) if 1.31 grams of silver bromide are produced?

Answer 1

Answer 2

Let's look at the balanced equation for the reaction taking place. It's always a good idea to do this.

AgNO₃(aq) + HBr(aq) ==> AgBr(s) + HNO₃

Next, we will find the theoretical yield of AgBr

moles of AgNO₃ = 25.0 ml x 1 L/1000 ml x 0.541 mol/L = 0.013525 moles AgNO₃

moles HBr = 10.0 mll x 1 L/1000 ml x 3.00 mol/L = 0.0300 moles HBr

Limiting reactant is AgNO₃

Theoretical yield of AgBr = 0.013525 mol AgNO₃ x 1 mol AgBr / mol AgNO₃ = 0.013525 mol AgBr

molar mass AgBr = 188 g/mol

Theoretical yield = 0.013525 mole AgBr x 188 g/mol = 2.54 g

% yield = actual yield/theoretical yield (x100) = 1.31 g/2.54 g (x100) = 51.6% yield

Answer 3

Hello! So there's a lot going on in the problem. My video solution breaks down everything you need to consider when solving this problem. A different tutor, J. R. S., gave the answer already, but the reasoning is explained here.

Hope this was helpful!

1. Matthew