Chapter 15, Problem 109A

a.

Interpretation Introduction

Interpretation: To determine the gas that is a limiting reactant.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The formation of products is dependent on the number of reactants. The reactant that is consumed first is known as the limiting reactant, when this reactant is consumed completely, then there will be no more formation of products.

Answer to Problem 109A

The limiting reactant is hydrogen gas.

Explanation of Solution

The amount of oxygen is $40{\text{ cm}}^3$ .

The amount of hydrogen is $60{\text{ cm}}^3$ .

In order to find the limiting reactant, calculate the number of moles of oxygen and hydrogen gas.

Calculation for moles of oxygen is calculated as follows:

  $40\cancel{{\text{cm}}^3}\times \frac{1\cancel{\text{mL}}}{1\cancel{{\text{cm}}^3}}\times \frac{1\cancel{\text{L}}}{1000\cancel{\text{mL}}}\times \frac{1{\text{ mol O}}_2}{22.4\cancel{\text{L}}}=1.8\times {10}^{-3}{\text{ mol O}}_2$

Calculation for moles of hydrogen is calculated as follows:

  $60\cancel{{\text{cm}}^3}\times \frac{1\cancel{\text{mL}}}{1\cancel{{\text{cm}}^3}}\times \frac{1\cancel{\text{L}}}{1000\cancel{\text{mL}}}\times \frac{1{\text{ mol H}}_2}{22.4\cancel{\text{L}}}=2.7\times {10}^{-3}{\text{ mol H}}_2$

  $2.7\times {10}^3\cancel{{\text{mol O}}_2}\times \frac{2{\text{ mol H}}_2}{1\cancel{{\text{mol O}}_2}}=3.6\times {10}^{-3}{\text{ mol H}}_2$

The moles of hydrogen is less. Thus, the limiting reagent is hydrogen.

b.

Interpretation Introduction

Interpretation: To calculate the mass of water.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction. The coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 109A

The amount of water is $0.049\text{ g}$ .

Explanation of Solution

The amount of water can be calculated as follows:

  $2.7\times {10}^{-3}\cancel{{\text{mol H}}_2}\times \frac{2\cancel{{\text{mol H}}_2\text{O}}}{2\cancel{{\text{mol H}}_2}}\times \frac{18.0{\text{ g H}}_2\text{O }}{1\cancel{{\text{mol H}}_2\text{O}}}=0.049{\text{ g H}}_2\text{O}$

c.

Interpretation Introduction

Interpretation: To give the name of the gas that is present after the reaction.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The formation of products is dependent on the number of reactants. The reactant that is consumed first is known as the limiting reactant, when this reactant is consumed completely, then there will be no more formation of products.

Answer to Problem 109A

The gas that is still present in the reaction is oxygen gas.

Explanation of Solution

The limiting reagent of this reaction is hydrogen gas which means oxygen gas is still present in the reaction and behaves as an excess reagent.

d.

Interpretation Introduction

Interpretation: To calculate the volume of gas at STP.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction. The coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 109A

The volume of the remaining gas at STP is $9\times {10}^{-3}\text{ L}$ .

Explanation of Solution

The number of moles of oxygen gas present can be calculated as follows:

  $2.7\times {10}^{-3}\cancel{{\text{mol H}}_2}\times \frac{1{\text{ mol O}}_2}{2\cancel{{\text{mol H}}_2}}=1.4\times {10}^{-3}{\text{ mol O}}_2$

Excess amount of oxygen can be given as:

  $\begin{array}{l}\left(1.8\times {10}^{-3}-1.4\times {10}^{-3}\right){\text{ mol O}}_2\\ =0.4\times {10}^{-3}{\text{ mol O}}_2\end{array}$

The volume of the remaining gas can be calculated as follows:

  $0.4\times {10}^{-3}\cancel{{\text{mol O}}_2}\times \frac{22.4{\text{ L O}}_2}{1\cancel{{\text{mol O}}_2}}=9\times {10}^{-3}\text{ L}$