A 55.0 mL sample of a 0.102 M potassium sulfate solution

Chapter 4, Problem 70E

(choose chapter or problem)

Get Unlimited Answers
QUESTION:

A 55.0 mL sample of a 0.102 M potassium sulfate solution is mixed with 35.0 mL of a 0.114 M lead(II) acetate solution and this precipitation reaction occurs:

The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

Questions & Answers

QUESTION:

A 55.0 mL sample of a 0.102 M potassium sulfate solution is mixed with 35.0 mL of a 0.114 M lead(II) acetate solution and this precipitation reaction occurs:

The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

ANSWER:

Solution: Here, we are going to identify which of the given pairs of solution has higher concentration of iodide ion. Step1: a) 1 molecule of BaI dissociates in aqueous solution to give 1 Ba ion and 2 I ions. 2+ - 2 - Therefore, 0.10 M BaI will g2e (2 x 0.10 = 0.20) M I ions in solution. Similarly, 1 molecule of KI dissociates in aqueous solution to give one K ion and one I ion. + - - Therefore, 0.25 M KI will produce 0.25 M I ions in solution. - Thus, 0.25 M KI solution has higher concentration of I ions. Step2: b) Number of moles of any com

Add to cart


Study Tools You Might Need

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back