Chapter 9, Problem 56A

(a)

Interpretation Introduction

Interpretation: The number of moles of products formed during the reaction of 1.00 g of ${\text{NH}}_{\text{3(g) }}$ with HCl needs to be determined as given in the reaction.

  ${\text{NH}}_{\text{3(g) }}{\text{+ HCl}}_{\text{(g)}}\stackrel{}{\to }{\text{ NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}$

Concept Introduction: A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant.

Answer to Problem 56A

  $\text{ 0}{\text{.059 moles NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}$

Explanation of Solution

The balance chemical equation:

  ${\text{NH}}_{\text{3(g) }}{\text{+ HCl}}_{\text{(g)}}\stackrel{}{\to }{\text{ NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}$

Mass of ${\text{NH}}_{\text{3(g) }}$ = 1.00 g

Molar mass of ${\text{NH}}_{\text{3(g) }}$ = 17.00 g

Calculate moles of ${\text{NH}}_{\text{3(g) }}$ :

  ${\text{Moles of NH}}_{\text{3(g) }}\text{= }\frac{\text{1}\text{.00}}{\text{17}\text{.00 g/mol}}\text{= 0}\text{.059 moles}$

From the balance chemical equation:

1 mole of ${\text{NH}}_{\text{3(g) }}$ = form 1 mole of ${\text{NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}$

Calculate moles of ${\text{NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}$ :

  $\text{0}{\text{.059 moles NH}}_{\text{3(g) }}\text{×}\frac{{\text{1 mole NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}}{{\text{1 mole NH}}_{\text{3(g) }}}\text{ = 0}{\text{.059 moles NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}$

(b)

Interpretation Introduction

Interpretation: The moles of products formed during the reaction of 1.00 g of $\text{CaO}$ with ${\text{CO}}_{\text{2}}$ as given in the reaction needs to be determined.

  ${\text{CaO}}_{\text{(s) }}{\text{+ CO}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ CaCO}}_{\text{3}}{}_{\text{(s)}}$

Concept Introduction: A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant.

Answer to Problem 56A

  $\text{0}{\text{.0178 moles CaCO}}_{\text{3}}$

Explanation of Solution

The balance chemical equation:

  ${\text{CaO}}_{\text{(s) }}{\text{+ CO}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ CaCO}}_{\text{3}}{}_{\text{(s)}}$

Mass of $\text{CaO}$ = 1.00 g

Molar mass of $\text{CaO}$ = 56.0 g

Calculate moles of $\text{CaO}$ :

  $\text{Moles of CaO= }\frac{\text{1}\text{.00}}{\text{56}\text{.00 g/mol}}\text{= 0}\text{.0178 moles}$

From the balance chemical equation:

1 mole of $\text{CaO}$ = form 1 mole of ${\text{ CaCO}}_{\text{3}}$

Calculate moles of ${\text{ CaCO}}_{\text{3}}$ :

  $\text{0}\text{.0178 moles CaO ×}\frac{{\text{1 mole CaCO}}_{\text{3}}}{\text{1 mole CaO}}\text{ = 0}{\text{.0178 moles CaCO}}_{\text{3}}$

(c)

Interpretation Introduction

Interpretation: The moles of products formed during the reaction of 1.00 g of $\text{Na}$ with ${\text{O}}_{\text{2}}$ as given in the reaction needs to be determined.

  ${\text{4 Na }}_{\text{(s) }}{\text{+ O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2 Na}}_{\text{2}}{\text{O}}_{\text{(s)}}$

Concept Introduction: A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant.

Answer to Problem 56A

  $\text{ 0}{\text{.0435 moles Na}}_{\text{2}}\text{O}$

Explanation of Solution

The balance chemical equation:

  ${\text{4 Na }}_{\text{(s) }}{\text{+ O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2 Na}}_{\text{2}}{\text{O}}_{\text{(s)}}$

Mass of $\text{Na}$ = 1.00 g

Molar mass of $\text{Na}$ = 23.0 g

Calculate moles of $\text{Na}$ :

  $\text{Moles of Na= }\frac{\text{1}\text{.00}}{\text{23}\text{.00 g/mol}}\text{= 0}\text{.0435 moles}$

From the balance chemical equation:

2 mole of $\text{Na}$ = form 1 mole of ${\text{Na}}_{\text{2}}\text{O}$

Calculate moles of ${\text{Na}}_{\text{2}}\text{O}$ :

  $\text{0}\text{.0435 moles Na ×}\frac{{\text{1 mole Na}}_{\text{2}}\text{O}}{\text{2 mole Na}}\text{ = 0}{\text{.02175 moles Na}}_{\text{2}}\text{O}$

(d)

Interpretation Introduction

Interpretation: The number of moles of products formed during the reaction of 1.00 g of $\text{P}$ with ${\text{Cl}}_{\text{2}}$ as given in the reaction needs to be determined.

  ${\text{2 P }}_{\text{(s) }}\text{+ 3}{\text{Cl}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2 PCl}}_{\text{3}}{}_{\text{(l)}}$

Concept Introduction: A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant.

Answer to Problem 56A

  $\text{0}{\text{.032 moles PCl}}_{\text{3}}$

Explanation of Solution

The balance chemical equation:

  ${\text{2 P }}_{\text{(s) }}\text{+ 3}{\text{Cl}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2 PCl}}_{\text{3}}{}_{\text{(l)}}$

Mass of $\text{P}$ = 1.00 g

Molar mass of $\text{P}$ = 30.9 g

Calculate moles of $\text{P}$ :

  $\text{Moles of P= }\frac{\text{1}\text{.00}}{\text{30}\text{.9 g/mol}}\text{= 0}\text{.032 moles}$

From the balance chemical equation:

1 mole of P = form 1 mole of ${\text{PCl}}_{\text{3}}$

Calculate moles of ${\text{PCl}}_{\text{3}}$ :

  $\text{0}\text{.032 moles P ×}\frac{{\text{1 mole PCl}}_{\text{3}}}{\text{1 mole P}}\text{ = 0}{\text{.032 moles PCl}}_{\text{3}}$