Ask a New Question

Solid lead(II) iodide was prepared by reacting 65.0 mL of a solution containing 0.218 M lead(II) ions with 80.0 mL of a solution containing 0.265 M iodide ions. If the actual yield of the reaction was 3.26 g, which choice is closest to the %yield of the reaction?

2 answers

and the choices are ....

is it 33.4%?

Ask a New Question or answer this question.

Similar Questions

We can precipitate out solid lead(II) iodide from an aqueous lead(II) nitrate solution by adding potassium iodide. If we have
1. 5 answers
When a lead nitrate solution is poured into a potassium iodide solution, a bright yellow solid called lead iodide forms.Which is
1. 1 answer
A solution containing excess lead(ii) nitrate is reacted with 380.0ML of 0.250mol/L potassium iodide solution. A bright yellow
1. 1 answer
One way to remove lead ion from water is to add a source of iodide ion so that lead iodide will precipitate out of solution:Pb2+
1. 2 answers

more similar questions