Chapter 3, Problem 108QRT

(a)

Interpretation Introduction

Interpretation:

The given ${\text{CaF}}_{\text{2}}{\text{(s) + H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq) }\to {\text{ HF(g) + CaSO}}_{\text{4}}\text{(s)}$ reaction has to be balanced and the name has to be given for the substrate.

Concept introduction:

Balancing equation:

Balanced equation is a chemical reaction in which number of atoms for each element in the reaction and the total charge are same on both reactant side and the product side.

Steps in balancing the equation

• Step 1: Write the unbalanced equation
• Step 2: Find the coefficient to balance the equation.
• The coefficient should be reduced to the smallest whole number.

Generally the ionic compound consists of cation and anion.

Cation: When electrons are removed from the atom that results to form positively charged ion called cation.

Anion: Addition of electron to atom results to form negatively charged ion called anion.

(b)

Interpretation Introduction

Interpretation:

The given ${\text{CaF}}_{\text{2}}{\text{(s) + H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq) }\to {\text{ HF(g) + CaSO}}_{\text{4}}\text{(s)}$ reaction is redox reactions, acid-base reactions or gas-forming reaction has to be identified.

Concept introduction:

Precipitation reaction: The formation of the product is insoluble when the ions combine in the solution is called precipitation reaction.

Acid - base reaction: Formation of the salt from the cation from the base and anion from the acid and formation of water is also the product.

Oxidation - reduction reaction: The electrons are transferred to one to other species is called oxidation reduction reaction.

(c)

Interpretation Introduction

Interpretation:

The compound name ${\text{CCl}}_{\text{4}}{\text{, SbCl}}_{\text{5}}\text{, and HCl}$ has to be given.

Concept introduction:

Refer to part (a)

(d)

Interpretation Introduction

Interpretation:

The elemental analysis report gives $8.74\%$ carbon, $\text{77}\text{.43\%}$ chlorine, and $\text{13}\text{.83\%}$ fluorine. The empirical formula of the compound has to be determined.

Concept introduction:

Number of moles can be calculated by using following formula,

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