Chapter 3, Problem 139QRT

(a)

Interpretation Introduction

Interpretation:

Based on the reaction, the product has to be predicted, name and formula for the product has to be given.

Explanation of Solution

Given,

Three Group 2A elements magnesium, calcium, and strontium are allowed to react with liquid bromine.

The product, name and formula for the product of the reaction is given below,

    The magnesium metal reaction with bromine produces magnesium bromide $\left({\text{MgBr}}_{\text{2}}\right)$ as a solid (s).

  The calcium metal reaction with bromine produces calcium bromide $\left({\text{CaBr}}_{\text{2}}\right)$ as a solid.

  The strontium metal reaction with bromine produces strontium bromide $\left({\text{SrBr}}_{\text{2}}\right)$ as a solid.

(b)

Interpretation Introduction

Interpretation:

The balanced equation has to be written for the given reaction.

Concept introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Balancing the equation:

• There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• First write the skeletal reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

Explanation of Solution

Given,

Three Group 2A elements magnesium, calcium, and strontium are allowed to react with liquid bromine.

The balanced equation is given below,

    The magnesium metal reaction with bromine produces magnesium bromide $\left({\text{MgBr}}_{\text{2}}\right)$ as a solid (s).

    $\text{Mg}{\text{(s) + Br}}_{\text{2}}\text{(l) }\to {\text{MgBr}}_{\text{2}}\text{(s)}$

  The calcium metal reaction with bromine produces calcium bromide $\left({\text{CaBr}}_{\text{2}}\right)$ as a solid.

  ${\text{Ca(s) + Br}}_{\text{2}}\text{(l) }\to {\text{CaBr}}_{\text{2}}\text{(s)}$

  The strontium metal reaction with bromine produces strontium bromide $\left({\text{SrBr}}_{\text{2}}\right)$ as a solid.

    ${\text{Sr(s) + Br}}_{\text{2}}\text{(l) }\to {\text{SrBr}}_{\text{2}}\text{(s)}$

(c)

Interpretation Introduction

Interpretation:

Nature of the reaction (acid-base reaction, precipitation reaction, or an oxidation-reduction reaction) has to be identified.

Concept introduction:

Precipitation reaction: The formation of the product is insoluble when the reactant 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.

Gas forming reaction: The reaction of acid and metal carbonates which produce carbonic acid. The carbonic acid decomposes which gives water and carbon dioxide.

Oxidation - reduction reaction: In this reaction, electrons are transferred from one to other reagent.

Explanation of Solution

Given,

Three Group 2A elements magnesium, calcium, and strontium are allowed to react with liquid bromine.

The magnesium metal reaction with bromine produces magnesium bromide $\left({\text{MgBr}}_{\text{2}}\right)$ as a solid.

    $\text{Mg}{\text{(s) + Br}}_{\text{2}}\text{(l) }\to {\text{MgBr}}_{\text{2}}\text{(s)}$

The given reaction is precipitation reaction, because the formation of ${\text{MgBr}}_{\text{2}}$ is not soluble in the water.

The calcium metal reaction with bromine produces calcium bromide $\left({\text{CaBr}}_{\text{2}}\right)$ as a solid.

  ${\text{Ca(s) + Br}}_{\text{2}}\text{(l) }\to {\text{CaBr}}_{\text{2}}\text{(s)}$

The given reaction is precipitation reaction, because the formation of ${\text{CaBr}}_{\text{2}}$ is not soluble in the water.

The strontium metal reaction with bromine produces strontium bromide $\left({\text{SrBr}}_{\text{2}}\right)$ as a solid.

    ${\text{Sr(s) + Br}}_{\text{2}}\text{(l) }\to {\text{SrBr}}_{\text{2}}\text{(s)}$

The given reaction is precipitation reaction, because the formation of ${\text{SrBr}}_{\text{2}}$ is not soluble in the water.

(d)

Interpretation Introduction

Interpretation:

Formula of each product has to be predicted.

Explanation of Solution

Given,

Figure 1

The mass of the metal is directly proportional to the mass of the compound formed.

From the graph, mass of sample is on the y-axis and the mass of the metal is on the x-axis and the fixed mass of the other reactant.

According to the graph (figure 1), mass of the compound is 11.4 g

Mass of the Mg (metal) is 1.6 g.

Consider the reaction takes place in 100 %. Then the mass of bromine is calculated as follows,

  $\begin{array}{l}Massofbro\min e=11.4gcompound-1.6gMg\\ Massofbro\min e=9.8g\end{array}$

The molar mass of Mg is 24.3050 g/mole, the molar mass of bromine is 79.904 g/mole.

Number of moles of Mg is calculated below,

  $\begin{array}{l}\text{ Moles =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{ Moles = }\frac{1.6\text{g}}{24.3050\text{g/mol}}\\ \text{ Moles = 0}\text{.066}\end{array}$

Number of moles of bromine is calculated below,

  $\begin{array}{l}\text{ Moles =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{ Moles = }\frac{9.8\text{g}}{\text{79}\text{.904}\text{g/mol}}\\ \text{ Moles = 0}\text{.123}\end{array}$

Therefore,

The ratio of the number of moles of Mg and bromine is given below,

  0.066: 0.123

Divide through by smallest number:

  $\begin{array}{l}\text{Mg = }\frac{0.066}{0.066}\\ \text{Mg = 1}\end{array}$

  $\begin{array}{l}\text{Br = }\frac{0.123}{0.066}\\ \text{Br = 1}\text{.9}\\ \text{Br = 2}\end{array}$

Therefore, the Empirical formula for the given compound is ${\text{Mg}}_{\text{1}}{\text{Br}}_{\text{2}}$. The Empirical formula for the given compound is calculated.

According to the graph (figure 1), mass of the compound is 12.4 g

Mass of the Ca (metal) is 2.5 g.

Consider the reaction takes place in 100 %. Then the mass of bromine is calculated as follows,

  $\begin{array}{l}Massofbro\min e=12.4gcompound-2.5gCa\\ Massofbro\min e=9.9g\end{array}$

The molar mass of Ca is 40.078 g/mole, the molar mass of bromine is 79.904 g/mole.

Number of moles of Ca is calculated below,

  $\begin{array}{l}\text{ Moles =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{ Moles = }\frac{2.5\text{g}}{40.078\text{g/mol}}\\ \text{ Moles = 0}\text{.062}\end{array}$

Number of moles of bromine is calculated below,

  $\begin{array}{l}\text{ Moles =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{ Moles = }\frac{9.9\text{g}}{\text{79}\text{.904}\text{g/mol}}\\ \text{ Moles = 0}\text{.124}\end{array}$

Therefore,

The ratio of the number of moles of Ca and bromine is given below,

  0.062: 0.124

Divide through by smallest number:

  $\begin{array}{l}\text{Ca = }\frac{0.062}{0.062}\\ \text{Ca =1}\end{array}$

  $\begin{array}{l}\text{Br = }\frac{0.124}{0.062}\\ \text{Br = 2}\end{array}$

Therefore, the Empirical formula for the given compound is ${\text{Ca}}_{\text{1}}{\text{Br}}_{\text{2}}$. The Empirical formula for the given compound is calculated.

According to the graph (figure 1), mass of the compound is 15.3 g

Mass of the Sr (metal) is 5.5 g.

Consider the reaction takes place in 100 %. Then the mass of bromine is calculated as follows,

  $\begin{array}{l}Massofbro\min e=15.3gcompound-5.5gCa\\ Massofbro\min e=9.8g\end{array}$

The molar mass of Sr is 87.62 g/mole, the molar mass of bromine is 79.904 g/mole.

Number of moles of Sr is calculated below,

  $\begin{array}{l}\text{ Moles =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{ Moles = }\frac{5.5\text{g}}{87.62\text{g/mol}}\\ \text{ Moles = 0}\text{.063}\end{array}$

Number of moles of bromine is calculated below,

  $\begin{array}{l}\text{ Moles =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{ Moles = }\frac{9.8\text{g}}{\text{79}\text{.904}\text{g/mol}}\\ \text{ Moles = 0}\text{.123}\end{array}$

Therefore,

The ratio of the number of moles of Sr and bromine is given below,

  0.063: 0.123

Divide through by smallest number:

  $\begin{array}{l}\text{Sr = }\frac{0.063}{0.063}\\ \text{Sr =1}\end{array}$

  $\begin{array}{l}\text{Br = }\frac{0.123}{0.063}\\ \text{Br = 1}\text{.9}\\ \text{Br = 2}\end{array}$

Therefore, the Empirical formula for the given compound is ${\text{Sr}}_{\text{1}}{\text{Br}}_{\text{2}}$. The Empirical formula for the given compound is calculated.