Chapter 9, Problem 43A

Interpretation Introduction

Interpretation:

The theoretical and percent yield of sodium thiosulfate pentahydrate needs to be deduced under the given reaction conditions.

Concept Introduction:

• 'Yield refers to the amount of product obtained from a reaction.
• 'Theoretical yield' refers to the amount of product obtained from the molar ratio in the balanced equation whereas the 'actual yield' represents the product recovered from an experimental procedure.
• 'Percent yield' is a ratio of the actual and theoretical yields.

  $\text{\% Yield = }\frac{\text{Actual yield (or experimental yield)}}{\text{Theoretical yield}}\text{×100 ------(1)}$

Answer to Problem 43A

% yield of sodium thiosulfate pentahydrate = 20.9%.

Explanation of Solution

Given:

Mass of S₈ = 3.25 g

Mass of Na₂SO₃ = 13.1 g

Actual yield of sodium thiosulfate pentahydrate=5.26 g

Calculations:

The given reaction is:

  ${\text{S}}_{\text{8}}{\text{ + Na}}_{\text{2}}{\text{SO}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O}\to {\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.5H}}_{\text{2}}\text{O}$

The balanced equation is:

  ${\text{S}}_{\text{8}}{\text{ + 8Na}}_{\text{2}}{\text{SO}}_{\text{3}}{\text{ + 40H}}_{\text{2}}\text{O}\to 8{\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.5H}}_{\text{2}}\text{O}$

Step 1: calculate the moles of S₈ as follows:

Mass of S₈ = 3.25 g

Molar mass of S₈ = 256.48 g/mol

  $\begin{array}{l}{\text{Moles of S}}_{\text{8}}=\frac{{\text{Mass of S}}_{\text{8}}}{{\text{Molar mass S}}_{\text{8}}}\\ =\frac{3.25\text{ g}}{256.48\text{ g/mol}}=0.0127\text{ moles}\end{array}$

Step 2: calculate the moles of Na₂SO₃ as follows:

Mass of Na₂SO₃ = 13.1 g

Molar mass of Na₂SO₃ = 126.04 g/mol

  $\begin{array}{l}{\text{Moles of Na}}_{\text{2}}{\text{SO}}_{\text{3}}=\frac{{\text{Mass of Na}}_{\text{2}}{\text{SO}}_{\text{3}}}{{\text{Molar mass Na}}_{\text{2}}{\text{SO}}_{\text{3}}}\\ =\frac{13.1\text{ g}}{126.04\text{ g/mol}}=0.1039\text{ moles}\end{array}$

Step 3: Deduce the limiting reagent as follows:

Based on the reaction stoichiometry:

1 mole of S₈ reacts with 8 moles of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$

Therefore, 0.0127 moles of S₈ will react with: $0.0127\times 8=0.1016{\text{ moles of Na}}_{\text{2}}{\text{SO}}_{\text{3}}$

From the calculations in step 2, there are 0.1039 moles of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ available, i.e., it is present in excess. Hence, S₈ is the limiting reagent and will determine the amount of product formed.

Step 4: Calculate the moles of ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.5H}}_{\text{2}}\text{O}$ formed as follows:

Based on the reaction stoichiometry:

1 mole of S₈ forms 8 moles of ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.5H}}_{\text{2}}\text{O}$

Therefore, 0.0127 moles of S8 will form:

  $\begin{array}{l}\frac{0.0127\text{ moles S8}\times {\text{ 8 moles Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.5H}}_{\text{2}}\text{O}}{1\text{ mole S8}}\\ =0.1016{\text{ moles Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.5H}}_{\text{2}}\text{O}\end{array}$

Step 5: Calculate the theoretical yield or amount of ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.5H}}_{\text{2}}\text{O}$ :

  ${\text{moles Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.5H}}_{\text{2}}\text{O}$ =0.1016

Molar mass ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.5H}}_{\text{2}}\text{O}$ = 248.18 g/mol

  $\begin{array}{l}{\text{Mass Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.5H}}_{\text{2}}\text{O = moles }\times \text{ molar mass}\\ \text{=0}\text{.1016}\times \text{248}\text{.18 = 25}\text{.21 g}\end{array}$

Step 6: Calculate the % yield

Based on equation (1):

  $\begin{array}{l}\text{\% Yield = }\frac{\text{Actual yield (or experimental yield)}}{\text{Theoretical yield}}\text{×100}\\ \text{=}\frac{5.26}{25.21}\times 100=20.9\%\end{array}$

Conclusion

Therefore, the yield of sodium thiosulfate pentahydrate is 20.9%.