Answered: All 4 problems please

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...

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All 4 problems please

Expert Solution

Step 1

“Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved, then please specify the question number or post only that question.”

The given chemical equation is:

$C a {(O H)}_{2} (a q) + 2 C O_{2} (g) \to C a {(H C O_{3})}_{2} (a q)$

From the given equation,

2 moles of CO₂ react with 1 mole Ca(OH)₂.

Step 2

(a) 6 moles of CO₂ will react with-

$\begin{array}{rcl} M o l e s o f C a {(O H)}_{2} & = & \frac{1 m o l C a {(O H)}_{2}}{2 m o l C O_{2}} \times 6 m o l C O_{2} \\ = & 3 m o l C a {(O H)}_{2} \end{array}$

5 moles of Ca(OH)₂ are given, CO₂ is the limiting reactant.

(b) Calculating moles of Ca(OH)₂:

$\begin{array}{rcl} M o l e s o f C a {(O H)}_{2} & = & \frac{5.24 g}{74.09 g / m o l} \\ = & 0.0707 m o l \end{array}$

Calculating moles of CO₂:

$\begin{array}{rcl} M o l e s o f C O_{2} & = & \frac{5.96 g}{44.01 g / m o l} \\ = & 0.135 m o l \end{array}$

0.135 moles of CO₂ will react with-

$\begin{array}{rcl} M o l e s o f C a {(O H)}_{2} & = & \frac{1 m o l C a {(O H)}_{2}}{2 m o l C O_{2}} \times 0.135 m o l C O_{2} \\ = & 0.0675 m o l C a {(O H)}_{2} \end{array}$

As the calculated moles of Ca(OH)₂ is 0.0707 moles. CO2 is the limiting reagent here.

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