Answered: When a solution containing 27.2 g of…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Concept explainers

Question

When a solution containing 27.2 g of lead (II) nitrate and a solution containing 35.4 g of potassium iodide are mixed, a yellow precipitate of lead (II) iodide is formed.

If 13.6 g of the precipitate formed, what was the percent yield for the reaction? Note: you must write out and balance the equation first.

Expert Solution

Step 1

The balanced reaction is:

$Pb {({NO}_{3})}_{2} (aq) + 2 KI (aq) \to {PbI}_{2} (s) + 2 {KNO}_{3} (aq)$

Calculation of no. of mol of Pb(NO₃)₂:

$\begin{array}{rcl} n & = & \frac{given mass}{Molar mass} \\ = & \frac{27.2 g}{331.2 g / mol} \\ = & 0.0821 mol \end{array}$

Calculation of no. of mol of KI:

$\begin{array}{rcl} n & = & \frac{given mass}{Molar mass} \\ = & \frac{35.4 g}{166 g / mol} \\ = & 0.21325 mol \end{array}$

Step by step

Solved in 4 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions