Chapter 3, Problem 3.145P

(a)

Interpretation Introduction

Interpretation:

The molar masses of the superconductor with a different value of x are to be calculated.

Concept introduction:

Molar mass is defined as the summation of the masses of the atoms that are present in one mole of a particular molecule.

A superconductor is a substance that conducts electricity without any resistance. The conductivity of these substances is very high.

Answer to Problem 3.145P

The molar masses of ${\text{La}}_{\text{2}}{\text{Sr}}_0{\text{CuO}}_{\text{4}}$ is $405.4\text{g/mol}$, ${\text{La}}_1{\text{Sr}}_1{\text{CuO}}_{\text{4}}$ is $354.1\text{g/mol}$, ${\text{La}}_{1.837}{\text{Sr}}_{0.163}{\text{CuO}}_{\text{4}}$ is $397\text{g/mol}$.

Explanation of Solution

When the value of x is zero the formula of the semiconductor is ${\text{La}}_{\text{2}}{\text{Sr}}_0{\text{CuO}}_{\text{4}}$. The formula to calculate the molar mass of ${\text{La}}_{\text{2}}{\text{Sr}}_0{\text{CuO}}_{\text{4}}$ is as follows:

  $\text{Molar mass of}{\text{La}}_{\text{2}}{\text{Sr}}_0{\text{CuO}}_{\text{4}}=\left[\begin{array}{l}2\left(\text{molar mass of La}\right)\\ +1\left(\text{molar mass of Cu}\right)\\ +4\left(\text{molar mass of O}\right)\end{array}\right]$        (1)

Substitute $138.9\text{g/mol}$ for molar mass of $\text{La}$, $63.55\text{g/mol}$ for molar mass of $\text{Cu}$ and $16.00\text{g/mol}$ for molar mass of $\text{O}$ in the equation (1).

  $\begin{array}{c}\text{Molar mass of}{\text{La}}_{\text{2}}{\text{Sr}}_0{\text{CuO}}_{\text{4}}=\left[\begin{array}{l}2\left(138.9\text{g/mol}\right)+1\left(63.55\text{g/mol}\right)\\ +4\left(16.00\text{g/mol}\right)\end{array}\right]\\ =405.4\text{g/mol}\end{array}$

When the value of x is 1 the formula of the semiconductor is ${\text{La}}_1{\text{Sr}}_1{\text{CuO}}_{\text{4}}$. The formula to calculate the molar mass of ${\text{La}}_1{\text{Sr}}_1{\text{CuO}}_{\text{4}}$ is as follows:

  $\text{Molar mass of}{\text{La}}_1{\text{Sr}}_1{\text{CuO}}_{\text{4}}=\left[\begin{array}{l}1\left(\text{molar mass of La}\right)\\ +1\left(\text{molar mass of Sr}\right)\\ +1\left(\text{molar mass of Cu}\right)\\ +4\left(\text{molar mass of O}\right)\end{array}\right]$        (2)

Substitute $138.9\text{g/mol}$ for molar mass of $\text{La}$, $87.62\text{g/mol}$ for molar mass of $\text{Sr}$ $63.55\text{g/mol}$ for molar mass of $\text{Cu}$ and $16.00\text{g/mol}$ for molar mass of $\text{O}$ in the equation (2).

  $\begin{array}{c}\text{Molar mass of}{\text{La}}_1{\text{Sr}}_1{\text{CuO}}_{\text{4}}=\left[\begin{array}{l}1\left(138.9\text{g/mol}\right)+1\left(87.62\text{g/mol}\right)\\ +1\left(63.55\text{g/mol}\right)+4\left(16.00\text{g/mol}\right)\end{array}\right]\\ =354.1\text{g/mol}\end{array}$

When the value of x is $0.163$ the formula of the semiconductor is ${\text{La}}_{1.837}{\text{Sr}}_{0.163}{\text{CuO}}_{\text{4}}$. The formula to calculate the molar mass of ${\text{La}}_{1.837}{\text{Sr}}_{0.163}{\text{CuO}}_{\text{4}}$ is as follows:

  $\text{Molar mass of}{\text{La}}_{1.837}{\text{Sr}}_{0.163}{\text{CuO}}_{\text{4}}=\left[\begin{array}{l}1.837\left(\text{molar mass of La}\right)\\ +0.163\left(\text{molar mass of Sr}\right)\\ +1\left(\text{molar mass of Cu}\right)\\ +4\left(\text{molar mass of O}\right)\end{array}\right]$        (3)

Substitute $138.9\text{g/mol}$ for molar mass of $\text{La}$, $87.62\text{g/mol}$ for molar mass of $\text{Sr}$ $63.55\text{g/mol}$ for molar mass of $\text{Cu}$ and $16.00\text{g/mol}$ for molar mass of $\text{O}$ in the equation (3).

  $\begin{array}{c}\text{Molar mass of}{\text{La}}_{1.837}{\text{Sr}}_{0.163}{\text{CuO}}_{\text{4}}=\left[\begin{array}{l}1.837\left(138.9\text{g/mol}\right)+0.163\left(87.62\text{g/mol}\right)\\ +1\left(63.55\text{g/mol}\right)+4\left(16.00\text{g/mol}\right)\end{array}\right]\\ =397\text{g/mol}\text{.}\end{array}$

Conclusion

${\text{La}}_{\text{2-x}}{\text{Sr}}_{\text{x}}{\text{CuO}}_{\text{4}}$ is a superconductor that conducts electricity without any resistance. The value of x changes the chemical formula as well as the mass of the semiconductor.

(b)

Interpretation Introduction

Interpretation:

The limiting reactant is to be identified.

Concept introduction:

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction.

Amount (mol) of excess reactant is reactant left after the formation of the maximum amount(mol) of products.

The formula to calculate moles is as follows:

  $\text{Amount}\left(\text{mol}\right)=\frac{\text{mass}}{\text{molar mass}}$

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

  $\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Answer to Problem 3.145P

${\text{BaCO}}_3$ is the limiting reactant.

Explanation of Solution

The chemical equation for the formation of a superconductor is written as follows:

  $\begin{array}{l}4{\text{BaCO}}_3\left(s\right)+6\text{CuO}\left(s\right)+{\text{Y}}_{\text{2}}{\text{O}}_3\left(s\right)\to 2{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_{6.5}\left(s\right)+4{\text{CO}}_2\left(g\right)\\ {\text{2YBa}}_2{\text{Cu}}_3{\text{O}}_{6.5}\left(s\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\to 2{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7\end{array}$

Consider the masses of ${\text{BaCO}}_3$, $\text{CuO}$ and ${\text{Y}}_{\text{2}}{\text{O}}_3$ to be $\text{x}\text{g}$.

The formula to calculate the moles of ${\text{BaCO}}_3$ is as follows:

  ${\text{Moles of BaCO}}_3=\left[\frac{\text{given mass}{\text{of BaCO}}_3}{{\text{molecular mass of BaCO}}_3}\right]$        (4)

Substitute $\text{x}\text{g}$ for the mass of ${\text{BaCO}}_3$ and $197.3\text{ g/mol}$ for the molecular mass of ${\text{BaCO}}_3$ in the equation (4).

  $\begin{array}{c}{\text{Moles of BaCO}}_3=\left[\frac{\text{x}\text{g}}{197.3\text{ g/mol}}\right]\\ =0.0050684\text{x}\text{mol}\end{array}$

The formula to calculate the moles of ${\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7$ from ${\text{BaCO}}_3$ is as follows:

  $\text{Moles of}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7=\text{moles of}{\text{BaCO}}_3\left(\frac{2\text{mol}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7}{4\text{mol}{\text{BaCO}}_3}\right)$        (5)

Substitute $0.0050684\text{x}\text{mol}$ for moles of ${\text{BaCO}}_3$ in the equation (5).

  $\begin{array}{c}\text{Moles of}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7=0.0050684\text{x}\text{mol}\left(\frac{2\text{mol}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7}{4\text{mol}{\text{BaCO}}_3}\right)\\ =0.002534\text{x}\text{mol}\end{array}$

The formula to calculate the moles of ${\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7$ from $\text{CuO}$ is as follows:

  $\text{Moles of}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7=\left(\frac{\text{mass of}\text{CuO}}{\text{molar mass of CuO}}\right)\left(\frac{2\text{mol}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7}{6\text{mol}\text{CuO}}\right)$        (6)

Substitute $\text{x}\text{g}$ for the mass of $\text{CuO}$ and $79.55\text{ g/mol}$ for the molar mass of $\text{CuO}$ in the equation (6).

  $\begin{array}{c}\text{Moles of}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7=\left(\frac{\text{x}\text{g}}{79.55\text{ g/mol}}\right)\left(\frac{2\text{mol}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7}{6\text{mol}\text{CuO}}\right)\\ =0.004190\text{x}\text{mol}\end{array}$

The formula to calculate the moles of ${\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7$ from ${\text{Y}}_{\text{2}}{\text{O}}_3$ is as follows:

  $\text{Moles of}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7=\left(\frac{\text{mass of}{\text{Y}}_{\text{2}}{\text{O}}_3}{{\text{molar mass of Y}}_{\text{2}}{\text{O}}_3}\right)\left(\frac{2\text{mol}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7}{1\text{mol}{\text{Y}}_{\text{2}}{\text{O}}_3}\right)$        (7)

Substitute $\text{x}\text{g}$ for the mass of ${\text{Y}}_{\text{2}}{\text{O}}_3$ and $225.82\text{ g/mol}$ for the molar mass of ${\text{Y}}_{\text{2}}{\text{O}}_3$ in the equation (7).

$\begin{array}{c}\text{Moles of}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7=\left(\frac{\text{x}\text{g}}{225.82\text{ g/mol}}\right)\left(\frac{2\text{mol}{\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7}{1\text{mol}{\text{Y}}_{\text{2}}{\text{O}}_3}\right)\\ =0.008857\text{x}\text{mol}\text{.}\end{array}$

${\text{BaCO}}_3$ is the limiting reactant, as it produces less amount of product ${\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7$.

Conclusion

The limiting reagent controls the amount of the other reactants and the product and produces less moles of product. ${\text{BaCO}}_3$ is the limiting reactant because the least moles of ${\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7$ is produced from ${\text{BaCO}}_3$.

(c)

Interpretation Introduction

Interpretation:

The mass percentage of each reactant in the remaining solid mixture is to be calculated.

Concept introduction:

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

  $\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Mass percent is employed to determine the concentration of one compound in a mixture of the compound. The formula to calculate mass percent is as follows:

  $\text{Mass percent of compound}=\left(\frac{\text{mass of the compound}}{\text{mass of mixture}}\right)100$

Answer to Problem 3.145P

The mass percentage of $\text{CuO}$ is $39.52\%$, the mass percentage of ${\text{Y}}_{\text{2}}{\text{O}}_3$ is $71.39\%$ and the mass percentage of ${\text{BaCO}}_3$ is zero.

Explanation of Solution

${\text{BaCO}}_3$ is the limiting reactant, as it produces less amount of product ${\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7$. The moles of ${\text{BaCO}}_3$ is zero.

Consider the masses of ${\text{BaCO}}_3$, $\text{CuO}$ and ${\text{Y}}_{\text{2}}{\text{O}}_3$ to be $\text{x}\text{g}$.

The formula to calculate the mass of $\text{CuO}$ remaining is as follows:

  $\text{Mass of}\text{CuO}=\left[\begin{array}{l}\left(\text{x}\text{g}\right)-\text{moles of}{\text{BaCO}}_3\text{reacted}\left(\frac{6\text{mol}\text{CuO}}{4\text{mol}{\text{BaCO}}_3}\right)\\ \left(\text{molar mass of }\text{CuO}\right)\end{array}\right]$        (8)

Substitute $0.0050684\text{x}\text{mol}$ for moles of ${\text{BaCO}}_3$ and $79.55\text{ g/mol}$ for the molar mass of $\text{CuO}$ in the equation (8).

  $\begin{array}{c}\text{Mass of}\text{CuO}=\left[\begin{array}{l}\left(\text{x}\text{g}\right)-\left(0.0050684\text{x}\text{mol}\right)\left(\frac{6\text{mol}\text{CuO}}{4\text{mol}{\text{BaCO}}_3}\right)\\ \left(79.55\text{ g/mol}\right)\end{array}\right]\\ =0.39521\text{x}\text{g}\end{array}$

The formula to calculate the mass of ${\text{Y}}_{\text{2}}{\text{O}}_3$ remaining is as follows:

  $\text{Mass of}{\text{Y}}_{\text{2}}{\text{O}}_3=\left[\begin{array}{l}\left(\text{x}\text{g}\right)-\text{moles of}{\text{BaCO}}_3\text{reacted}\left(\frac{1\text{mol}{\text{Y}}_{\text{2}}{\text{O}}_3}{4\text{mol}{\text{BaCO}}_3}\right)\\ \left(\text{molar mass of}{\text{Y}}_{\text{2}}{\text{O}}_3\right)\end{array}\right]$        (9)

Substitute $0.0050684\text{x}\text{mol}$ for moles of ${\text{BaCO}}_3$ and $225.82\text{ g/mol}$ for the molar mass of ${\text{Y}}_{\text{2}}{\text{O}}_3$ in the equation (9).

$\begin{array}{c}\text{Mass of}{\text{Y}}_{\text{2}}{\text{O}}_3=\left[\begin{array}{l}\left(\text{x}\text{g}\right)-\left(0.0050684\text{x}\text{mol}\right)\left(\frac{1\text{mol}{\text{Y}}_{\text{2}}{\text{O}}_3}{4\text{mol}{\text{BaCO}}_3}\right)\\ \left(225.82\text{ g/mol}\right)\end{array}\right]\\ =0.713862\text{x}\text{g}\end{array}$

The formula to calculate the mass percentage of $\text{CuO}$ is as follows:

  $\text{Mass percent of CuO}=\left(\frac{\text{mass of CuO}}{\text{Initial mass}}\right)100$        (10)

Substitute $0.39521\text{x}\text{g}$ for the mass of $\text{CuO}$ and $\text{x}\text{g}$ for the initial mass in the equation (10).

  $\begin{array}{c}\text{Mass percent of CuO}=\left(\frac{0.39521\text{x}\text{g}}{\text{x}\text{g}}\right)100\\ =39.521\%\\ \approx 39.52\%\end{array}$

The formula to calculate the mass percentage of ${\text{Y}}_{\text{2}}{\text{O}}_3$ is as follows:

  ${\text{Mass percent of Y}}_{\text{2}}{\text{O}}_3=\left(\frac{{\text{mass of Y}}_{\text{2}}{\text{O}}_3}{\text{Initial mass}}\right)100$        (11)

Substitute $0.713862\text{x}\text{g}$ for the mass of ${\text{Y}}_{\text{2}}{\text{O}}_3$ and $\text{x}\text{g}$ for the initial mass in the equation (11).

  $\begin{array}{c}{\text{Mass percent of Y}}_{\text{2}}{\text{O}}_3=\left(\frac{0.713862\text{x}\text{g}}{\text{x}\text{g}}\right)100\\ =71.3862\%\\ \approx 71.39\%.\end{array}$

Conclusion

${\text{BaCO}}_3$ is the limiting reactant because the least moles of ${\text{YBa}}_2{\text{Cu}}_3{\text{O}}_7$ is produced from ${\text{BaCO}}_3$. ${\text{BaCO}}_3$ is consumed in the reaction and therefore the mass percentage of ${\text{BaCO}}_3$ is zero.