Chapter 3, Problem 94QRT

(a)

Interpretation Introduction

Interpretation:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}{\text{(aq) + Br}}_{\text{2}}\text{(aq) }\to {\text{ 2 HBr(aq) + C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}\text{(aq)}$

$1.00g$ Chewable vitamin C tablet requires $27.85mLof0.102MB{r}_2$ to reacts completely.

The step for the mass in grams of vitamin C in the tablet is incorrect has to be identified.

Concept introduction:

Number of moles can be calculated by using following formula,

  $\text{No}\text{.of Moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}$

The Molarity of the solution can be calculated by using following formula

  $\text{Molarity of the solution}\text{=}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litre}}$

Explanation of Solution

The reaction is shown below,

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}{\text{(aq) + Br}}_{\text{2}}\text{(aq) }\to {\text{ 2 HBr(aq) + C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}\text{(aq)}$

$1.00g$ Chewable vitamin C tablet requires $27.85mLof0.102MB{r}_2$ to reacts completely.

According to the balanced equation, 1 moles of citric acid (${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}$) reaction with one moles of ${\text{Br}}_{\text{2}}$ gives two mole of hydrogen bromide and one moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}$. Therefore,

Step 1:

Given,

  ${\text{n (Br}}_{\text{2}}\text{) = (27}\text{.85 mL)(0}\text{.102 mol/L)}$

Moles of ${\text{Br}}_{\text{2}}$ is calculated as follows,

    $\begin{array}{l}\text{Moles}\text{of}{\text{Br}}_{\text{2}}\text{=}\text{27}\text{.85 mL×}\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\text{×}\frac{\text{0}\text{.102 mol}}{\text{1L}}\\ \text{Moles}\text{of}{\text{Br}}_{\text{2}}\text{=}\text{2}{\text{.84×10}}^{\text{-3}}\text{mol}\end{array}$

Step 1 is incorrect answer, because the conversion of mL to litter is missing.

Step 2:

Given,

  ${\text{n (C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{) = }\left(\text{2}{\text{.84 mol Br}}_{\text{2}}\right)\text{ × }\left(\frac{{\text{1 mol C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}}{{\text{1 mol Br}}_{\text{2}}}\right)$

Moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}$ is calculated as follows,

    $\begin{array}{l}\text{Moles}{\text{of C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{ = }\left(\text{2}{\text{.84×10}}^{\text{-3}}\text{mol}\right)\text{ × }\left(\frac{{\text{1 mol C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}}{{\text{1 mol Br}}_{\text{2}}}\right)\\ \text{Moles}{\text{of C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{ = 2}{\text{.84×10}}^{\text{-3}}\text{mol}\end{array}$

Step is 2 is incorrect because, number of moles of bromine is wrong. Hence Step is 2 is incorrect.

Step 3:

Given,

  ${\text{m (C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{) = (2}{\text{.84 mol C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{)}\times {\text{(176 g/mol C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{)}$

Moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}$ is $\text{2}{\text{.84×10}}^{\text{-3}}\text{mol}$

    $\begin{array}{l}{\text{m (C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{) = 2}{\text{.84×10}}^{\text{-3}}\text{mol}\times \frac{\text{176}{\text{.1238 g C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}}{{\text{1 mol C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}}\\ {\text{m (C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{) = 0}{\text{.5 g C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\end{array}$

Step is 3 is incorrect because, moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}$ is wrong. Hence Step is 3 is incorrect.

Step 4:

Given,

  ${\text{m (C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{) = }\frac{\text{(500}{\text{. g C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{)}}{\text{(1 g tablet)}}$

Step is 4 is incorrect because, mass in grams of ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}$ is too large, hence it is wrong. Therefore, Step is 4 is incorrect.

Over all, step 1, step 2, step 3 and step 4 are incorrect answer.

(b)

Interpretation Introduction

Interpretation:

The correct result has to be determined.

Concept introduction:

Refer to part (a)

Explanation of Solution

The reaction is shown below,

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}{\text{(aq) + Br}}_{\text{2}}\text{(aq) }\to {\text{ 2 HBr(aq) + C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}\text{(aq)}$

$1.00g$ Chewable vitamin C tablet requires $27.85mLof0.102MB{r}_2$ to reacts completely.

According to the balanced equation, 1 moles of citric acid (${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}$) reaction with one moles of ${\text{Br}}_{\text{2}}$ gives two mole of hydrogen bromide and one moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}$. Therefore,

Moles of ${\text{Br}}_{\text{2}}$ is calculated as follows,

    $\begin{array}{l}\text{Moles}\text{of}{\text{Br}}_{\text{2}}\text{=}\text{27}\text{.85 mL×}\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\text{×}\frac{\text{0}\text{.102 mol}}{\text{1L}}\\ \text{Moles}\text{of}{\text{Br}}_{\text{2}}\text{=}\text{2}{\text{.84×10}}^{\text{-3}}\text{mol}\end{array}$

Moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}$ is calculated as follows,

    $\begin{array}{l}\text{Moles}{\text{of C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{ = }\left(\text{2}{\text{.84×10}}^{\text{-3}}\text{mol}\right)\text{ × }\left(\frac{{\text{1 mol C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}}{{\text{1 mol Br}}_{\text{2}}}\right)\\ \text{Moles}{\text{of C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{ = 2}{\text{.84×10}}^{\text{-3}}\text{mol}\end{array}$

The mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}$ is calculated as follows,

    $\begin{array}{l}{\text{m (C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{) = 2}{\text{.84×10}}^{\text{-3}}\text{mol}\times \frac{\text{176}{\text{.1238 g C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}}{{\text{1 mol C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}}\\ {\text{m (C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\text{) = 0}{\text{.5 g C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}\end{array}$

The correct result is that the mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}$ is $\text{0}\text{.5 g}$.