Chapter 9, Problem 27A

Interpretation Introduction

Interpretation:

The amount of chlorine gas required to produce 5.00x10-3 g of chlorine mono fluoride in an excess of fluorine gas needs to be deduced based on the given reaction.

Concept Introduction:

• A chemical reaction is represented in terms of a chemical equation with the reactants on the left and the products on the right.

  $\text{Reactants }\to \text{ Products}$

• The coefficient of a balanced chemical equation i.e. the stoichiometry gives the amount of reactants and products involved in the reaction.
• Chemical equations can therefore be used to determine the amount of products formed from a known quantity of reactants.

Answer to Problem 27A

Mass of Cl₂ required = 3.25x10-3 g

Explanation of Solution

The given reaction is:

  ${\text{Cl}}_{\text{2}}{\text{(g) + F}}_2\text{(g) }\to \text{2ClF(g)}$

Step 1: Calculate the moles of ClF formed:

Mass of ClF formed = 5.00x10-3 g

Molar mass of ClF = 54.45 g/mol

  $\begin{array}{l}\text{Moles of ClF = }\frac{\text{Mass of ClF}}{\text{Molar Mass ClF}}\\ =\frac{\text{5}\text{.0}\times {\text{10}}^{-3}\text{ g}}{54.45\text{ g/mol}}=\text{0}\text{.0918}\times {\text{10}}^{-3}\text{ moles}\end{array}$

Step 2: Calculate the moles of Cl₂ needed:

Based on the reaction stoichiometry:

1 mole of Cl₂ forms 2 mole of ClF

Therefore, 0.0918x10-3 moles of ClF will be formed upon reaction with:

  $\frac{1{\text{ mole Cl}}_2\times \text{0}\text{.0918}\times {\text{10}}^{-3}\text{ moles ClF}}{2\text{ moles ClF}}=0.0459\times {\text{10}}^{-3}{\text{ moles Cl}}_{\text{2}}$

Step 3: Calculate the mass of Cl₂ needed:

Moles of Cl₂ needed= 0.0459x10-3

Molecular weight of Cl₂ = 70.9 g/mol

  $\begin{array}{l}{\text{Mass of Cl}}_2\text{ = Moles }\times \text{ Molecular weight}\\ =0.0459\times {10}^{-3}\text{ moles }\times \text{ 70}\text{.9 g/mol = 3}\text{.25}\times {\text{10}}^{-3}\text{ g}\end{array}$

Conclusion

Therefore, mass of Cl₂ required is around 3.25x10-3 g.