Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 3, Problem 3.111QP

(a)

Interpretation Introduction

Interpretation: The theoretical yield of the products of each of the given unbalanced chemical equations for the given reactions is to be determined.

Concept introduction: The mass of product formed in the given reaction depends upon the mass of the limiting reagent in the reaction.

To determine: The mass of product formed in the given unbalanced reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 3.111QP

Solution

The mass of Li3N formed in the given reaction is 4.97g .

Explanation of Solution

Explanation

Given

The unbalanced chemical equation for the given reaction is,

Li(s)+N2(g)Li3N(s)

The mass of lithium in the reaction is 5.0g .

The mass of nitrogen gas in the reaction is 2.0g .

The given reaction contains one lithium atom and two nitrogen atoms on the reactant side and three lithium atoms and one nitrogen atom on the product side. Therefore, the given reaction is balanced by multiplying the Li with 6 and Li3N with 2 .

6Li(s)+N2(g)2Li3N(s)

The number of moles of the given element is calculated by using the formula,

Numberofmoles=MassMolarmass (1)

Substitute the values of mass and molar mass of lithium in the equation (1).

MolesofLi=5.00g6.94g/mol=0.7204mol

Substitute the values of mass and molar mass of nitrogen in the equation (1).

MolesofN2=2.00g28.014g/mol=0.0714mol

The number of moles of nitrogen required in the given reaction is,

MolesofN2=16×0.7204mol(Li)=0.120mol

That is more than the number of moles of nitrogen present in the reaction medium. Therefore, nitrogen is the limiting reagent in the given reaction.

Since, the number of moles of N2(g) is equal to the half of the moles of Li3N in the given reaction. Therefore, the number of moles of Li3N formed in the given reaction is double as compared to N2(g) as the reaction stops immediately after the concentration of N2(g) finishes.

The number of moles of Li3N formed is 2×0.0714mol=0.1428mol .

The mass of Li3N formed in the given reaction is calculated by multiplying its number of moles with the molar mass.

MassofLi3N=0.1428mol×34.827g/mol=4.97g

Thus, the mass of Li3N formed in the given reaction is 4.97g .

(b)

Interpretation Introduction

To determine: The mass of product formed in the given unbalanced reaction.

(b)

Expert Solution
Check Mark

Answer to Problem 3.111QP

Solution

The mass of H3PO4 formed in the given reaction is 34.5g .

Explanation of Solution

Explanation

Given

The unbalanced chemical equation for the given reaction is,

P2O5(s)+H2O(l)H3PO4(aq)

The mass of P2O5 in the reaction is 25.0g .

The mass of water in the reaction is 36.0g .

The given reaction contains two phosphorus atoms, six oxygen atoms and two hydrogen atoms on the reactant side and three hydrogen atoms, one phosphorus atom and four oxygen atoms on the product side. Therefore, the given reaction is balanced by multiplying the H2O with 3 and H3PO4 with 2 .

P2O5(s)+3H2O(l)2H3PO4(aq)

The number of moles of the given element is calculated by using the formula,

Numberofmoles=MassMolarmass (1)

Substitute the values of mass and molar mass of P2O5 in the equation (1).

MolesofP2O5=25.0g141.941g/mol=0.1761mol

Substitute the values of mass and molar mass of H2O in the equation (1).

MolesofH2O=36.0g18.02g/mol=1.998mol

The number of moles of P2O5 required in the given reaction is,

MolesofP2O5=13×1.998mol(H2O)=0.666mol

That is more than the number of moles of P2O5 present in the reaction medium. Therefore, P2O5 is the limiting reagent in the given reaction.

Since, the number of moles of P2O5 is equal to the half of the moles of H3PO4 in the given reaction. Therefore, the number of moles of H3PO4 formed in the given reaction is double as compared to P2O5 as the reaction stops immediately after the concentration of P2O5 finishes.

The number of moles of H3PO4 formed is 2×0.1761mol=0.3522mol .

The mass of H3PO4 formed in the given reaction is calculated by multiplying its number of moles with the molar mass.

MassofH3PO4=0.3522mol×97.993g/mol=34.5g

Thus, the mass of H3PO4 formed in the given reaction is 34.5g .

(c)

Interpretation Introduction

To determine: The mass of product formed in the given unbalanced reaction.

(c)

Expert Solution
Check Mark

Answer to Problem 3.111QP

Solution

The mass of SO3 formed in the given reaction is 7.93g .

Explanation of Solution

Explanation

Given

The unbalanced chemical equation for the given reaction is,

SO2(g)+O2(g)SO3(g)

The mass of SO2 in the reaction is 6.4g .

The mass of O2 in the reaction is 4.0g .

The given reaction contains one sulfur atom and four oxygen atoms on the reactant side and one sulfur atom and three oxygen atoms on the product side. Therefore, the given reaction is balanced by multiplying the SO2 with 2 and SO3 with 2 .

2SO2(g)+O2(g)2SO3(g)

The number of moles of the given element is calculated by using the formula,

Numberofmoles=MassMolarmass (1)

Substitute the values of mass and molar mass of SO2 in the equation (1).

MolesofSO2=6.4g64.058g/mol=0.099mol

Substitute the values of mass and molar mass of O2 in the equation (1).

MolesofO2=4.0g31.998g/mol=0.125mol

The number of moles of SO2 required in the given reaction is,

MolesofSO2=2×0.125mol(O2)=0.25mol

That is more than the number of moles of SO2 present in the reaction medium. Therefore, SO2 is the limiting reagent in the given reaction.

Since, the number of moles of SO2 is equal to the number of moles of SO3 in the given reaction. Therefore, the number of moles of SO2 formed in the given reaction is equal to that of SO2 as the reaction stops immediately after the concentration of SO2 finishes.

The number of moles of SO3 formed is 0.099mol .

The mass of SO3 formed in the given reaction is calculated by multiplying its number of moles with the molar mass.

MassofSO3=0.099mol×80.057g/mol=7.93g

Thus, the mass of SO3 formed in the given reaction is 7.93g .

Conclusion

  1. a. The mass of Li3N formed in the given reaction is 4.97g .
  2. b. The mass of H3PO4 formed in the given reaction is 34.5g .
  3. c. The mass of SO3 formed in the given reaction is 7.93g

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Chapter 3 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 3.4 - Prob. 11PECh. 3.5 - Prob. 12PECh. 3.5 - Prob. 13PECh. 3.6 - Prob. 14PECh. 3.6 - Prob. 15PECh. 3.6 - Prob. 16PECh. 3.7 - Prob. 17PECh. 3.8 - Prob. 18PECh. 3.8 - Prob. 19PECh. 3.9 - Prob. 20PECh. 3.9 - Prob. 21PECh. 3.9 - Prob. 22PECh. 3.9 - Prob. 23PECh. 3 - Prob. 3.1VPCh. 3 - Prob. 3.2VPCh. 3 - Prob. 3.3VPCh. 3 - Prob. 3.4VPCh. 3 - Prob. 3.5VPCh. 3 - Prob. 3.6VPCh. 3 - Prob. 3.7VPCh. 3 - Prob. 3.8VPCh. 3 - Prob. 3.9VPCh. 3 - Prob. 3.10VPCh. 3 - Prob. 3.11QPCh. 3 - Prob. 3.12QPCh. 3 - Prob. 3.13QPCh. 3 - Prob. 3.14QPCh. 3 - Prob. 3.15QPCh. 3 - Prob. 3.16QPCh. 3 - Prob. 3.17QPCh. 3 - Prob. 3.18QPCh. 3 - Prob. 3.19QPCh. 3 - Prob. 3.20QPCh. 3 - Prob. 3.21QPCh. 3 - Prob. 3.22QPCh. 3 - Prob. 3.23QPCh. 3 - Prob. 3.24QPCh. 3 - Prob. 3.25QPCh. 3 - Prob. 3.26QPCh. 3 - Prob. 3.27QPCh. 3 - Prob. 3.28QPCh. 3 - Prob. 3.29QPCh. 3 - Prob. 3.30QPCh. 3 - Prob. 3.31QPCh. 3 - Prob. 3.32QPCh. 3 - Prob. 3.33QPCh. 3 - Prob. 3.34QPCh. 3 - Prob. 3.35QPCh. 3 - Prob. 3.36QPCh. 3 - Prob. 3.37QPCh. 3 - Prob. 3.38QPCh. 3 - Prob. 3.39QPCh. 3 - Prob. 3.40QPCh. 3 - Prob. 3.41QPCh. 3 - Prob. 3.42QPCh. 3 - Prob. 3.43QPCh. 3 - Prob. 3.44QPCh. 3 - Prob. 3.45QPCh. 3 - Prob. 3.46QPCh. 3 - Prob. 3.47QPCh. 3 - Prob. 3.48QPCh. 3 - Prob. 3.49QPCh. 3 - Prob. 3.50QPCh. 3 - Prob. 3.51QPCh. 3 - Prob. 3.52QPCh. 3 - Prob. 3.53QPCh. 3 - Prob. 3.54QPCh. 3 - Prob. 3.55QPCh. 3 - Prob. 3.56QPCh. 3 - Prob. 3.57QPCh. 3 - Prob. 3.58QPCh. 3 - Prob. 3.59QPCh. 3 - Prob. 3.60QPCh. 3 - Prob. 3.61QPCh. 3 - Prob. 3.62QPCh. 3 - Prob. 3.63QPCh. 3 - Prob. 3.64QPCh. 3 - Prob. 3.65QPCh. 3 - Prob. 3.66QPCh. 3 - Prob. 3.67QPCh. 3 - Prob. 3.68QPCh. 3 - Prob. 3.69QPCh. 3 - Prob. 3.70QPCh. 3 - Prob. 3.71QPCh. 3 - Prob. 3.72QPCh. 3 - Prob. 3.73QPCh. 3 - Prob. 3.74QPCh. 3 - Prob. 3.75QPCh. 3 - Prob. 3.76QPCh. 3 - Prob. 3.77QPCh. 3 - Prob. 3.78QPCh. 3 - Prob. 3.79QPCh. 3 - Prob. 3.80QPCh. 3 - Prob. 3.81QPCh. 3 - Prob. 3.82QPCh. 3 - Prob. 3.83QPCh. 3 - Prob. 3.84QPCh. 3 - Prob. 3.85QPCh. 3 - Prob. 3.86QPCh. 3 - Prob. 3.87QPCh. 3 - Prob. 3.88QPCh. 3 - Prob. 3.89QPCh. 3 - Prob. 3.90QPCh. 3 - Prob. 3.91QPCh. 3 - Prob. 3.92QPCh. 3 - Prob. 3.93QPCh. 3 - Prob. 3.94QPCh. 3 - Prob. 3.95QPCh. 3 - Prob. 3.96QPCh. 3 - Prob. 3.97QPCh. 3 - Prob. 3.98QPCh. 3 - Prob. 3.99QPCh. 3 - Prob. 3.100QPCh. 3 - Prob. 3.101QPCh. 3 - Prob. 3.102QPCh. 3 - Prob. 3.103QPCh. 3 - Prob. 3.104QPCh. 3 - Prob. 3.105QPCh. 3 - Prob. 3.106QPCh. 3 - Prob. 3.107QPCh. 3 - Prob. 3.108QPCh. 3 - Prob. 3.109QPCh. 3 - Prob. 3.110QPCh. 3 - Prob. 3.111QPCh. 3 - Prob. 3.112QPCh. 3 - Prob. 3.113QPCh. 3 - Prob. 3.114QPCh. 3 - Prob. 3.115QPCh. 3 - Prob. 3.116QPCh. 3 - Prob. 3.117QPCh. 3 - Prob. 3.118QPCh. 3 - Prob. 3.119QPCh. 3 - Prob. 3.120QPCh. 3 - Prob. 3.121APCh. 3 - Prob. 3.122APCh. 3 - Prob. 3.123APCh. 3 - Prob. 3.124APCh. 3 - Prob. 3.125APCh. 3 - Prob. 3.126APCh. 3 - Prob. 3.127APCh. 3 - Prob. 3.128APCh. 3 - Prob. 3.129APCh. 3 - Prob. 3.130APCh. 3 - Prob. 3.131APCh. 3 - Prob. 3.132APCh. 3 - Prob. 3.133APCh. 3 - Prob. 3.134APCh. 3 - Prob. 3.135APCh. 3 - Prob. 3.136APCh. 3 - Prob. 3.137APCh. 3 - Prob. 3.138APCh. 3 - Prob. 3.139APCh. 3 - Prob. 3.140APCh. 3 - Prob. 3.141APCh. 3 - Prob. 3.142APCh. 3 - Prob. 3.143APCh. 3 - Prob. 3.144APCh. 3 - Prob. 3.145APCh. 3 - Prob. 3.146APCh. 3 - Prob. 3.147APCh. 3 - Prob. 3.148AP
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