CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
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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.

(a)

Expert Solution
Check Mark

Answer to Problem 3.145P

The molar masses of La2Sr0CuO4 is 405.4g/mol, La1Sr1CuO4 is 354.1g/mol, La1.837Sr0.163CuO4 is 397g/mol.

Explanation of Solution

When the value of x is zero the formula of the semiconductor is La2Sr0CuO4. The formula to calculate the molar mass of La2Sr0CuO4 is as follows:

  Molar mass ofLa2Sr0CuO4=[2(molar mass of La)+1(molar mass of Cu)+4(molar mass of O)]        (1)

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

  Molar mass ofLa2Sr0CuO4=[2(138.9g/mol)+1(63.55g/mol)+4(16.00g/mol)]=405.4g/mol

When the value of x is 1 the formula of the semiconductor is La1Sr1CuO4. The formula to calculate the molar mass of La1Sr1CuO4 is as follows:

  Molar mass ofLa1Sr1CuO4=[1(molar mass of La)+1(molar mass of Sr)+1(molar mass of Cu)+4(molar mass of O)]        (2)

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

  Molar mass ofLa1Sr1CuO4=[1(138.9g/mol)+1(87.62g/mol)+1(63.55g/mol)+4(16.00g/mol)]=354.1g/mol

When the value of x is 0.163 the formula of the semiconductor is La1.837Sr0.163CuO4. The formula to calculate the molar mass of La1.837Sr0.163CuO4 is as follows:

  Molar mass ofLa1.837Sr0.163CuO4=[1.837(molar mass of La)+0.163(molar mass of Sr)+1(molar mass of Cu)+4(molar mass of O)]        (3)

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

  Molar mass ofLa1.837Sr0.163CuO4=[1.837(138.9g/mol)+0.163(87.62g/mol)+1(63.55g/mol)+4(16.00g/mol)]=397g/mol.

Conclusion

La2-xSrxCuO4 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:

  Amount(mol)=massmolar 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,

  A+2B3C

One mole of A reacts with two moles of B to produce three moles of 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.

(b)

Expert Solution
Check Mark

Answer to Problem 3.145P

BaCO3 is the limiting reactant.

Explanation of Solution

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

  4BaCO3(s)+6CuO(s)+Y2O3(s)2YBa2Cu3O6.5(s)+4CO2(g)2YBa2Cu3O6.5(s)+12O2(g)2YBa2Cu3O7

Consider the masses of BaCO3, CuO and Y2O3 to be xg.

The formula to calculate the moles of BaCO3 is as follows:

  Moles of BaCO3=[given massof BaCO3molecular mass of BaCO3]        (4)

Substitute xg for the mass of BaCO3 and 197.3 g/mol for the molecular mass of BaCO3 in the equation (4).

  Moles of BaCO3=[xg197.3 g/mol]=0.0050684xmol

The formula to calculate the moles of YBa2Cu3O7 from BaCO3 is as follows:

  Moles ofYBa2Cu3O7=moles ofBaCO3(2molYBa2Cu3O74molBaCO3)        (5)

Substitute 0.0050684xmol for moles of BaCO3 in the equation (5).

  Moles ofYBa2Cu3O7=0.0050684xmol(2molYBa2Cu3O74molBaCO3)=0.002534xmol

The formula to calculate the moles of YBa2Cu3O7 from CuO is as follows:

  Moles ofYBa2Cu3O7=(mass ofCuOmolar mass of CuO)(2molYBa2Cu3O76molCuO)        (6)

Substitute xg for the mass of CuO and 79.55 g/mol for the molar mass of CuO in the equation (6).

  Moles ofYBa2Cu3O7=(xg79.55 g/mol)(2molYBa2Cu3O76molCuO)=0.004190xmol

The formula to calculate the moles of YBa2Cu3O7 from Y2O3 is as follows:

  Moles ofYBa2Cu3O7=(mass ofY2O3molar mass of Y2O3)(2molYBa2Cu3O71molY2O3)        (7)

Substitute xg for the mass of Y2O3 and 225.82 g/mol for the molar mass of Y2O3 in the equation (7).

Moles ofYBa2Cu3O7=(xg225.82 g/mol)(2molYBa2Cu3O71molY2O3)=0.008857xmol.

BaCO3 is the limiting reactant, as it produces less amount of product YBa2Cu3O7.

Conclusion

The limiting reagent controls the amount of the other reactants and the product and produces less moles of product. BaCO3 is the limiting reactant because the least moles of YBa2Cu3O7 is produced from BaCO3.

(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,

  A+2B3C

One mole of A reacts with two moles of B to produce three moles of 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:

  Mass percent of compound=(mass of the compoundmass of mixture)100

(c)

Expert Solution
Check Mark

Answer to Problem 3.145P

The mass percentage of CuO is 39.52%, the mass percentage of Y2O3 is 71.39% and the mass percentage of BaCO3 is zero.

Explanation of Solution

BaCO3 is the limiting reactant, as it produces less amount of product YBa2Cu3O7. The moles of BaCO3 is zero.

Consider the masses of BaCO3, CuO and Y2O3 to be xg.

The formula to calculate the mass of CuO remaining is as follows:

  Mass ofCuO=[(xg)moles ofBaCO3reacted(6molCuO4molBaCO3)(molar mass of CuO)]        (8)

Substitute 0.0050684xmol for moles of BaCO3 and 79.55 g/mol for the molar mass of CuO in the equation (8).

  Mass ofCuO=[(xg)(0.0050684xmol)(6molCuO4molBaCO3)(79.55 g/mol)]=0.39521xg

The formula to calculate the mass of Y2O3 remaining is as follows:

  Mass ofY2O3=[(xg)moles ofBaCO3reacted(1molY2O34molBaCO3)(molar mass ofY2O3)]        (9)

Substitute 0.0050684xmol for moles of BaCO3 and 225.82 g/mol for the molar mass of Y2O3 in the equation (9).

Mass ofY2O3=[(xg)(0.0050684xmol)(1molY2O34molBaCO3)(225.82 g/mol)]=0.713862xg

The formula to calculate the mass percentage of CuO is as follows:

  Mass percent of CuO=(mass of CuOInitial mass)100        (10)

Substitute 0.39521xg for the mass of CuO and xg for the initial mass in the equation (10).

  Mass percent of CuO=(0.39521xgxg)100=39.521%39.52%

The formula to calculate the mass percentage of Y2O3 is as follows:

  Mass percent of Y2O3=(mass of Y2O3Initial mass)100        (11)

Substitute 0.713862xg for the mass of Y2O3 and xg for the initial mass in the equation (11).

  Mass percent of Y2O3=(0.713862xgxg)100=71.3862%71.39%.

Conclusion

BaCO3 is the limiting reactant because the least moles of YBa2Cu3O7 is produced from BaCO3. BaCO3 is consumed in the reaction and therefore the mass percentage of BaCO3 is zero.

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Chapter 3 Solutions

CHEMISTRY >CUSTOM<

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