How many mL of o.677m HBrias needed tocamputely of Naz cOg?nectralize 0.8989Nan co, +2 HBr-> Ha COz + 2 NaBr-7-33

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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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How many mL of o.677m HBrias needed to
camputely of Naz cOg?
nectralize 0.8989
Nan co, +2 HBr-> Ha COz + 2 NaBr
-7-
3
3

Expert Solution

Step 1

The balanced equation for the complete neutralization reaction between ${Na}_{2} {CO}_{3} and HBr$ is as follows.

${Na}_{2} {CO}_{3} + 2 HBr \to H_{2} {CO}_{3} + 2 {Na}_{2} Br$

That means we need ${Na}_{2} {CO}_{3} and HBr$ in $1 : 2$ ratio for complete neutralization.

Step 2

The no of moles of ${Na}_{2} {CO}_{3}$ available can be calculated as follows.

$\begin{array}{rcl} no of moles & = & \frac{mass of {Na}_{2} {CO}_{3}}{molar mass of {Na}_{2} {CO}_{3}} \\ = & \frac{0.898 g}{105.97 g / mol} \\ = & 8.474 \times 10^{- 3} mol \end{array}$

The no of moles of $HBr$ needed to react with $8 \begin{array}{rcl} . \end{array} \begin{array}{rcl} 474 \end{array} \begin{array}{rcl} \times \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} ^{- 3} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} mol \end{array} \begin{array}{rcl} {Na}_{2} {CO}_{3} is \end{array}$ $\begin{array}{rcl} (2 \times 8.474 \times 10^{- 3}) & = & 0.01695 mol \end{array}$ .

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