Chapter 9, Problem 36A

(a)

Interpretation Introduction

Interpretation:

The limiting reactant and mass of each product in the given reaction needs to be determined.

  $S\left(s\right)+H_{2}S{O}_4\left(aq\right)\to S{O}_2\left(g\right)+H_{2}O\left(l\right)$

Concept Introduction:

The balanced equation is:

  $\begin{array}{l}\text{S}\left(\text{s}\right){\text{ + 2H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\to {\text{ 3SO}}_{\text{2}}\left(\text{g}\right){\text{ + 2H}}_{\text{2}}\text{O}\left(\text{l}\right)\hfill \\ \text{1 mol 2 mol 3 mol 2 mol}\hfill \\ \hfill \end{array}$

Answer to Problem 36A

  $S\left(s\right)+H_{2}S{O}_4\left(aq\right)\to S{O}_2\left(g\right)+H_{2}O\left(l\right)$

H₂SO₄ is the limiting reactant

mass of SO₂= 4.90 g SO₂

mass of H₂O = 0.919 g H₂O

Explanation of Solution

  $\begin{array}{l}\text{S}\left(\text{s}\right){\text{ + 2H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\to {\text{ 3SO}}_{\text{2}}\left(\text{g}\right){\text{ + 2H}}_{\text{2}}\text{O}\left(\text{l}\right)\hfill \\ \text{1 mol 2 mol 3 mol 2 mol}\hfill \\ \hfill \end{array}$

The given mass of each reactant is 5.00 g

mass of S(s) = 5.00 g

mass of H₂SO₄ = 5.00 g

To determine the limiting reactant, first calculate the mass of H₂SO₄ needed for complete consumption of 5.00 g of S(s).

  ${\text{mass of H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{= 5}\text{.00 g of S × }\left(\frac{\text{1 mol S}}{\text{32}\text{.06 g S}}\right)\text{×}\left(\frac{{\text{2 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1 mol S}}\right)\text{×}\left(\frac{\text{98}{\text{.08 g H}}_{\text{2}}{\text{SO}}_{\text{4}}}{{\text{1 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\text{ = 30}{\text{.6 g H}}_{\text{2}}{\text{SO}}_{\text{4}}$

The given amount of H₂SO₄ is 5.00 g, therefore, H₂SO₄ is the limiting reactant, because it is completely consumed in the reaction.

Use the quantity of the limiting reactant, H₂SO₄, to calculate the mass of each product expected.

  $\begin{array}{l}{\text{mass of SO}}_{\text{2}}\text{= 5}{\text{.00 g of H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{× }\left(\frac{{\text{ 1mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{98}{\text{.08 g H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\text{×}\left(\frac{{\text{3 mol SO}}_{\text{2}}}{{\text{2 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\text{×}\left(\frac{\text{64}{\text{.07 g SO}}_{\text{2}}}{{\text{1 mol SO}}_{\text{2}}}\right)\text{ = 4}{\text{.90 g SO}}_{\text{2}}\hfill \\ \hfill \\ {\text{mass of H}}_{\text{2}}\text{O = 5}{\text{.00 g of H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{× }\left(\frac{{\text{1 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{98}{\text{.08 g H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\text{×}\left(\frac{{\text{2molH}}_{\text{2}}\text{O}}{{\text{2 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\text{×}\left(\frac{\text{18}{\text{.02 g H}}_{\text{2}}\text{O}}{{\text{1 mol H}}_{\text{2}}\text{O}}\right)\text{ = 0}{\text{.919 g H}}_{\text{2}}\text{O}\hfill \\ \hfill \end{array}$

(b)

Interpretation Introduction

Interpretation:

The limiting reactant and mass of each product in the given reaction needs to be determined.

  $Mn{O}_2\left(s\right)+H_{2}S{O}_4\left(l\right)\to Mn{\left(S{O}_4\right)}_2\left(s\right)+H_{2}O\left(l\right)$

Concept Introduction:

The balanced equation is:

  $\begin{array}{l}{\text{MnO}}_{\text{2}}\left(\text{s}\right){\text{ + 2H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{l}\right)\to \text{Mn}{\left({\text{SO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right){\text{ + 2H}}_{\text{2}}\text{O}\left(\text{l}\right)\hfill \\ \text{1 mol 2 mol 1 mol 2 mol}\hfill \end{array}$

Answer to Problem 36A

  $Mn{O}_2\left(s\right)+H_{2}S{O}_4\left(l\right)\to Mn{\left(S{O}_4\right)}_2\left(s\right)+H_{2}O\left(l\right)$

H₂SO₄ is the limiting reactant

mass of Mn(SO₄)₂= 6.30 g Mn(SO₄)₂

mass of H₂O = 0.919 g H₂O

.

Explanation of Solution

  $Mn{O}_2\left(s\right)+H_{2}S{O}_4\left(l\right)\to Mn{\left(S{O}_4\right)}_2\left(s\right)+H_{2}O\left(l\right)$

The balanced equation is:

  $\begin{array}{l}{\text{MnO}}_{\text{2}}\left(\text{s}\right){\text{ + 2H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{l}\right)\to \text{ Mn}{\left({\text{SO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right){\text{ + 2H}}_{\text{2}}\text{O}\left(\text{l}\right)\hfill \\ \text{1 mol 2 mol 1 mol 2 mol}\hfill \end{array}$

The given mass of each reactant is 5.00 g

mass of MnO₂ = 5.00 g

mass of H₂SO₄ = 5.00 g

To determine the limiting reactant, we first calculate the mass of H₂SO₄ needed for complete consumption of 5.00 g of MnO₂.

  $\begin{array}{l}{\text{mass of H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{= 5}{\text{.00 g of MnO}}_{\text{2}}\text{×}\left(\frac{{\text{1 mol MnO}}_{\text{2}}}{\text{86}{\text{.94 g MnO}}_{\text{2}}}\right)\text{×}\left(\frac{{\text{2 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}{{\text{1 mol MnO}}_{\text{2}}}\right)\text{×}\left(\frac{\text{98}{\text{.08 g H}}_{\text{2}}{\text{SO}}_{\text{4}}}{{\text{1 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\text{ = 11}{\text{.3 g H}}_{\text{2}}{\text{SO}}_{\text{4}}\hfill \\ \hfill \end{array}$

The given amount of H₂SO₄ is 5.00 g, therefore, H₂SO₄ is the limiting reactant, because it is completely consumed in the reaction.

Use the quantity of the limiting reactant, H₂SO₄, to calculate the mass of each product expected.

  $\begin{array}{l}\text{mass of Mn}{\left({\text{SO}}_{\text{4}}\right)}_{\text{2}}\text{= 5}{\text{.00 g of H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{× }\left(\frac{{\text{1 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{98}{\text{.08 g H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\text{×}\left(\frac{\text{1 mol Mn}{\left({\text{SO}}_{\text{4}}\right)}_{\text{2}}}{{\text{2 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\text{×}\left(\frac{\text{247}\text{.06 g Mn}{\left({\text{SO}}_{\text{4}}\right)}_{\text{2}}}{\text{1 mol Mn}{\left({\text{SO}}_{\text{4}}\right)}_{\text{2}}}\right)\text{ = 6}\text{.30 g Mn}{\left({\text{SO}}_{\text{4}}\right)}_{\text{2}}\hfill \\ \hfill \\ {\text{mass of H}}_{\text{2}}\text{O = 5}{\text{.00 g of H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{× }\left(\frac{{\text{1 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{98}{\text{.08 g H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\text{×}\left(\frac{{\text{2 mol H}}_{\text{2}}\text{O}}{{\text{2 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\text{×}\left(\frac{\text{18}{\text{.02 g H}}_{\text{2}}\text{O}}{{\text{1 mol H}}_{\text{2}}\text{O}}\right)\text{ = 0}{\text{.919 g H}}_{\text{2}}\text{O}\hfill \end{array}$

(c)

Interpretation Introduction

Interpretation:

The limiting reactant and mass of each product in the given reaction needs to be determined.

  $H_{2}S\left(g\right)+O_2\left(g\right)\to S{O}_2\left(g\right)+H_{2}O\left(l\right)$

Concept Introduction:

The balanced equation is:

  $\begin{array}{l}{\text{2H}}_{\text{2}}\text{S}\left(\text{g}\right){\text{ + 3O}}_{\text{2}}\left(\text{g}\right)\to {\text{ 2SO}}_{\text{2}}\left(\text{g}\right){\text{ + 2H}}_{\text{2}}\text{O}\left(\text{l}\right)\hfill \\ \text{2 mol 3 mol 2 mol 2 mol}\hfill \end{array}$

Answer to Problem 36A

  $H_{2}S\left(g\right)+O_2\left(g\right)\to S{O}_2\left(g\right)+H_{2}O\left(l\right)$

O₂ is the limiting reactant

mass of SO₂= 6.67 g SO₂

mass of H₂O = 1.88 g H₂O

Explanation of Solution

  $H_{2}S\left(g\right)+O_2\left(g\right)\to S{O}_2\left(g\right)+H_{2}O\left(l\right)$

The balanced equation is:

  $\begin{array}{l}{\text{2H}}_{\text{2}}\text{S}\left(\text{g}\right){\text{ + 3O}}_{\text{2}}\left(\text{g}\right)\to {\text{ 2SO}}_{\text{2}}\left(\text{g}\right){\text{ + 2H}}_{\text{2}}\text{O}\left(\text{l}\right)\hfill \\ \text{2 mol 3 mol 2 mol 2 mol}\hfill \end{array}$

The given mass of each reactant is 5.00 g

mass of H₂S = 5.00 g

mass of O₂ = 5.00 g

To determine the limiting reactant, we first calculate the mass of O₂ needed for complete consumption of 5.00 g of H₂S

  ${\text{mass of O}}_{\text{2}}\text{= 5}{\text{.00 g of H}}_{\text{2}}\text{S ×}\left(\frac{{\text{1 mol H}}_{\text{2}}\text{S}}{\text{34}{\text{.1 g H}}_{\text{2}}\text{S}}\right)\left(\frac{{\text{3 mol O}}_{\text{2}}}{{\text{2 mol H}}_{\text{2}}\text{S}}\right)\left(\frac{\text{32}{\text{.00 g O}}_{\text{2}}}{{\text{1 mol O}}_{\text{2}}}\right)\text{ = 7}{\text{.04 g O}}_{\text{2}}$

The given amount of O₂ is 5.00 g, therefore, O₂ is the limiting reactant, because it is completely consumed in the reaction.

Use the quantity of the limiting reactant, O₂, to calculate the mass of each product expected.

  $\begin{array}{l}{\text{mass of SO}}_{\text{2}}\text{= 5}{\text{.00 g of O}}_{\text{2}}\text{× }\left(\frac{{\text{1 mol O}}_{\text{2}}}{\text{32}{\text{.00 g O}}_{\text{2}}}\right)\left(\frac{{\text{2 mol SO}}_{\text{2}}}{{\text{3 mol O}}_{\text{2}}}\right)\left(\frac{\text{64}{\text{.07 g SO}}_{\text{2}}}{{\text{1 mol SO}}_{\text{2}}}\right)\text{ = 6}{\text{.67 g SO}}_{\text{2}}\hfill \\ \hfill \\ {\text{mass of H}}_{\text{2}}\text{O = 5}{\text{.00 g of O}}_{\text{2}}\text{× }\left(\frac{{\text{1 mol O}}_{\text{2}}}{\text{32}{\text{.00 g O}}_{\text{2}}}\right)\left(\frac{{\text{2 mol H}}_{\text{2}}\text{O}}{{\text{3 mol O}}_{\text{2}}}\right)\left(\frac{\text{18}{\text{.02 g H}}_{\text{2}}\text{O}}{{\text{1 mol H}}_{\text{2}}\text{O}}\right)\text{ = 1}{\text{.88 g H}}_{\text{2}}\text{O}\hfill \\ \hfill \end{array}$

(d)

Interpretation Introduction

Interpretation:

The limiting reactant and mass of each product in the given reaction needs to be determined.

  $AgN{O}_3\left(aq\right)+Al\left(s\right)\to Ag\left(s\right)+Al{\left(N{O}_3\right)}_3\left(aq\right)$

Concept Introduction:

The balanced equation is:

  $\begin{array}{l}{\text{3AgNO}}_{\text{3}}\left(\text{aq}\right)\text{ + Al}\left(\text{s}\right)\to \text{ 3Ag}\left(\text{s}\right)\text{ + Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\left(\text{aq}\right)\hfill \\ \text{3 mol 1 mol 3 mol 1 mol}\hfill \\ \hfill \end{array}$

Answer to Problem 36A

  $AgN{O}_3\left(aq\right)+Al\left(s\right)\to Ag\left(s\right)+Al{\left(N{O}_3\right)}_3\left(aq\right)$

limiting reactant is AgNO₃.

mass of Ag= 3.18 g Ag

mass of Al(NO₃)₃ = 2.09 g Al(NO₃)₃

Explanation of Solution

  $AgN{O}_3\left(aq\right)+Al\left(s\right)\to Ag\left(s\right)+Al{\left(N{O}_3\right)}_3\left(aq\right)$

The balanced equation is:

  $\begin{array}{l}{\text{3AgNO}}_{\text{3}}\left(\text{aq}\right)\text{ + Al}\left(\text{s}\right)\to \text{ 3Ag}\left(\text{s}\right)\text{ + Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\left(\text{aq}\right)\hfill \\ \text{3 mol 1 mol 3 mol 1 mol}\hfill \\ \hfill \end{array}$

The given mass of each reactant is 5.00 g

mass of AgNO₃ = 5.00 g

mass of Al = 5.00 g

To determine the limiting reactant, we first calculate the mass of Al(s) needed for complete consumption of 5.00 g of AgNO₃

  $\begin{array}{l}\text{mass of Al = 5}{\text{.00 g of AgNO}}_{\text{3}}\text{× }\left(\frac{{\text{1 mol AgNO}}_{\text{3}}}{\text{169}{\text{.87 g AgNO}}_{\text{3}}}\right)\left(\frac{\text{1 mol Al}}{{\text{3 mol AgNO}}_{\text{3}}}\right)\left(\frac{\text{26}\text{.98 g Al}}{\text{1 mol Al}}\right)\\ \text{= 0}\text{.265 g Al}\end{array}$

The given amount of Al is 5.00 g, therefore, Al is in excess. The limiting reactant is AgNO₃, completely consumed in the reaction.

Use the quantity of the limiting reactant, AgNO₃, to calculate the mass of each product expected.

  $\begin{array}{l}\begin{array}{l}\begin{array}{l}\text{mass of Ag= 5}{\text{.00 g of AgNO}}_{\text{3}}\left(\frac{{\text{1 mol AgNO}}_{\text{3}}}{\text{169}{\text{.87 g AgNO}}_{\text{3}}}\right)\left(\frac{\text{3 mol Ag}}{{\text{3 mol AgNO}}_{\text{3}}}\right)\left(\frac{\text{107}\text{.9 g Ag}}{\text{1 mol Ag}}\right)\\ \text{ = 3}\text{.18 g Ag}\end{array}\hfill \\ \hfill \\ \hfill \end{array}\\ \begin{array}{l}\begin{array}{l}\text{mass of Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\text{= 5}{\text{.00 g of AgNO}}_{\text{3}}\left(\frac{{\text{1 mol AgNO}}_{\text{3}}}{\text{169}{\text{.87 g AgNO}}_{\text{3}}}\right)\left(\frac{\text{1 mol Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}}{{\text{3 mol AgNO}}_{\text{3}}}\right)\left(\frac{\text{213}\text{.0 g Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}}{\text{1 mol Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}}\right)\\ \text{ = 2}\text{.09 g Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\end{array}\hfill \\ \hfill \end{array}\end{array}$