Chapter 9, Problem 35A

(a)

Interpretation Introduction

Interpretation:

The reactant which is limiting needs to be determined and the mass of NaNH₂ which is expected needs to be determined.

  $2N{H}_3\left(g\right)+2Na\left(s\right)\to 2NaN{H}_2\left(s\right)+H_2\left(g\right)$

Concept Introduction:

  $2N{H}_3\left(g\right)+2Na\left(s\right)\to 2NaN{H}_2\left(s\right)+H_2\left(g\right)$

Calculate mol using the following expression:

  $\text{number of moles = }\frac{mass}{molar\text{ mass}}$

Answer to Problem 35A

The limiting reagent is Na

mass NaNH₂= 84.63 g

Explanation of Solution

Given data:

mass NH₃= 50 .0 g

mass Na = 50.0 g

  $2N{H}_3\left(g\right)+2Na\left(s\right)\to 2NaN{H}_2\left(s\right)+H_2\left(g\right)$

Now, calculate mol using the following expression:

  $\text{number of moles = }\frac{mass}{molar\text{ mass}}$

Molar Mass NH₃ = 17 g/mol

  ${\text{number of moles of NH}}_3\text{=}\frac{50.0g}{17.0g/mol}$

mol NH₃ = 2.94 mol

Molar Mass Na = 23.0 g/mol

  $\text{number of moles of Na=}\frac{50.0g}{23.0g/mol}$

mol Na = 2.17 mol

Now, calculate limiting reagent for this it is assumed that all the Na is completely consumed the NH3 mol necessary for all the Na to react completely 

  $\begin{array}{l}2N{H}_3\left(g\right)+2Na\left(s\right)\to 2NaN{H}_2\left(s\right)+H_2\left(g\right)\hfill \\ \begin{array}{l}X2.17mol\\ mol\text{ Na=}\frac{2.17molNa\times 2molN{H}_3}{2molNa}\end{array}\hfill \end{array}$

mol Na = 2.17 mol

note that the NH₃ moles needed for all Na to react completely is less than the amount initially available so Na is the limiting reagent and NH₃ is the excess reagent 

Now by stoichiometry we calculate mol NaNH₂

  $\begin{array}{l}2N{H}_3\left(g\right)+2Na\left(s\right)\to 2NaN{H}_2\left(s\right)+H_2\left(g\right)\hfill \\ \begin{array}{l}X2.17mol\\ mol\text{ Na=}\frac{2.17molNa\times 2mol\text{ Na}N{H}_2}{2molNa}\end{array}\hfill \end{array}$

mol NaNH₂= 2.17 mol

Now we calculate mass NaNH₂ using the following expression :

  $\begin{array}{l}mass=number\text{ of moles}\times \text{Molar mass}\\ {\text{mass of NaNH}}_2=2.17mol\times 39g/mol\\ =84.63g\end{array}$

(b)

Interpretation Introduction

Interpretation:

The reactant which is limiting needs to be determined and the mass of BaSO₄which is expected needs to be determined.

  $BaC{l}_2\left(aq\right)+N{a}_{2}S{O}_4\left(aq\right)\to BaS{O}_4\left(s\right)+2NaCl\left(aq\right)$

Concept Introduction:

Calculate mol using the following expression

  $\text{number of moles = }\frac{mass}{molar\text{ mass}}$

Molar Mass BaCl₂= 208.2 g/mol

Answer to Problem 35A

The limiting reagent is BaCl₂

mass BaSO₄= 56.01 g

Explanation of Solution

mass BaCl₂ = 50.0 g

mass NaSO₄ = 50.0 g

We calculate mol using the following expression

  $\text{number of moles = }\frac{mass}{molar\text{ mass}}$

Molar Mass BaCl₂= 208.2 g/mol

  ${\text{number of moles of BaCl}}_2\text{= }\frac{50.0g}{208.2g/mol}$

mol BaCl₂= 0.240 mol

Molar Mass Na₂SO₄= 142.06 g/mol

  ${\text{number of moles of Na}}_2{\text{SO}}_4\text{= }\frac{50.0g}{142.06g/mol}$

mol Na₂SO₄= 0.352 mol

  $BaC{l}_2\left(aq\right)+N{a}_{2}S{O}_4\left(aq\right)\to BaS{O}_4\left(s\right)+2NaCl\left(aq\right)$

Now, calculate limiting reagent for this it is assumed that all the BaCl₂ is completely consumed, for the Na₂SO₄ mol necessary for all the Na to react completely 

  $\begin{array}{l}\begin{array}{l}BaC{l}_2\left(aq\right)+N{a}_{2}S{O}_4\left(aq\right)\to BaS{O}_4\left(s\right)+2NaCl\left(aq\right)\hfill \\ 0.240molX\hfill \end{array}\\ molN{a}_{2}S{O}_4=\frac{0.240mol\times 1molN{a}_{2\text{}}S{O}_4\text{}\text{}}{1molBaC{l}_2\text{}}\end{array}$

mol Na₂SO₄= 0.240 mol

note that the Na₂SO₄ moles needed for all BaCl₂ to react completely is less than the amount initially available so BaCl2 is the limiting reagent and Na₂SO₄ is the excess reagent 

Now by stoichiometry we calculate mol BaSO₄

  $\begin{array}{l}\begin{array}{l}BaC{l}_2\left(aq\right)+N{a}_{2}S{O}_4\left(aq\right)\to BaS{O}_4\left(s\right)+2NaCl\left(aq\right)\hfill \\ 0.240molX\hfill \end{array}\\ mol\text{ Ba}S{O}_4=\frac{0.240molBaC{l}_2\times 1molBaS{O}_4\text{}\text{}}{1molBaC{l}_2\text{}}\end{array}$

mol BaSO4= 0.240 mol

Now, calculate mass BaSO₄

  $\begin{array}{l}mass=number\text{ of moles}\times \text{Molar mass}\\ \text{mass of Ba}S{O}_4=0.240mol\times 233.38g/mol\\ =56.01g\end{array}$

(c)

Interpretation Introduction

Interpretation:

The reactant which is limiting needs to be determined and the mass of Na₂SO₃which is expected needs to be determined.

SO₂(g) + 2 NaOH(aq) → Na₂SO₃(s) + H₂O(l)

Concept Introduction:

Calculate mol using the following expression

  $\text{number of moles = }\frac{mass}{molar\text{ mass}}$

Molar Mass SO₂ = 64.06 g/mol

Answer to Problem 35A

The limiting reagent is NaOH

Mass Na₂SO₃= 78.8 g

Explanation of Solution

mass SO2 = 50.0 g

mass NaOH = 50.0 g

Calculate mol using the following expression

  $\text{number of moles = }\frac{mass}{molar\text{ mass}}$

Molar Mass SO₂ = 64.06 g/mol

  $molS{O}_2=\frac{50.0g}{64.06g/mol}\text{}$

mol SO₂= 0.781 mol

Molar Mass NaOH = 50.0 g/mol

  $molNaOH=\frac{50.0g\text{}}{40.0g/mol}$

mol NaOH = 1.25 mol

SO₂(g) + 2 NaOH(aq) → Na₂SO₃(s) + H₂O(l)

Calculate limiting reagent for this it is assumed that all the NaOH is completely consumed for the SO2 mol necessary for all the NaOH to react completely 

Calculate

  $\begin{array}{l}S{O}_2\left(g\right)+2NaOH\left(aq\right)\to N{a}_{2}S{O}_3\left(s\right)+H_{2}O\left(l\right)\\ X1.25mol \\ molSO2=\frac{1.25mol\text{ NaOH}\times 1molS{O}_2}{2molNaOH}\text{}\end{array}$

mol SO2= 0.625 mol

note that the SO2 moles needed for all NaOH to react completely is less than the amount initially available.sSo, NaOH is the limiting reagent and SO2 is the excess reagent 

by stoichiometry we calculate mol Na₂SO₃ from limiting reagent

  $\begin{array}{l}S{O}_2\left(g\right)+2NaOH\left(aq\right)\to N{a}_{2}S{O}_3\left(s\right)+H_{2}O\left(l\right)\\ X1.25mol \\ molSO2=\frac{1.25molNaOH\times 1molN{a}_{2}S{O}_3}{2molNaOH}\text{}\end{array}$

mol Na₂SO₃= 0.625 mol

Molar mass Na₂SO₃= 126.06 g/mol

  $\begin{array}{l}mass=number\text{ of moles}\times \text{Molar mass}\\ {\text{mass of Na}}_{2}S{O}_4=0.625mol\times 126.06g/mol\\ =78.8g\end{array}$

(d)

Interpretation Introduction

Interpretation:

The limiting reagent needs to be determined. Also, the mass of Al₂(SO₄)₃needs to be determined.

  $2Al\left(s\right)+3H_{2}S{O}_4\left(aq\right)\to A{l}_2{(S{O}_4)}_3\left(s\right)+3H_2\left(g\right)$

Concept Introduction:

The number of moles can be calculated using the following expression

  $\text{number of moles = }\frac{mass}{molar\text{ mass}}$

Answer to Problem 35A

limiting reagent is H₂SO₄

mass Al₂(SO₄)₃ = 58.17 g

Explanation of Solution

mass Al = 50.0 g/mol

mass H₂SO₄ = 50.0 g/mol

First, calculate mol using the following expression

  $\text{number of moles = }\frac{mass}{molar\text{ mass}}$

  $\text{number of moles of Al=}\frac{50.0g}{27g/mol}$

mol Al = 1.85 mol

Molar H₂SO₄ = 98 g/mol

  ${\text{number of moles of H}}_2{\text{SO}}_4\text{=}\frac{50.0g}{98g/mol}$

mol H₂SO₄ = 0.510 mol

  $2Al\left(s\right)+3H_{2}S{O}_4\left(aq\right)\to A{l}_2{(S{O}_4)}_3\left(s\right)+3H_2\left(g\right)$

calculate limiting reagent for this it is assumed that all the H₂SO₄ is completely consumed for the mol of Al necessary for all the H₂SO₄ to react completely 

  $\begin{array}{l}2Al\left(s\right)+3H_{2}S{O}_4\left(aq\right)\to A{l}_2{(}_S\left(s\right)+3H_2\left(g\right)\\ X0.510mol\\ molAl=\text{}\frac{0.510mol{H}_{2\text{}}S{O}_4\text{}\times 2molAl}{3mol{H}_2\text{}S{O}_4}\text{}\end{array}$

mol Al = 0.340 mol

note that the Al moles needed for all H₂SO₄ to react completely is less than the amount initially available so H₂SO₄ is the limiting reagent and Al is the excess reagent 

Now from limiting reagent we calculate mol Al₂(SO₄)₃

  $\begin{array}{l}2Al\left(s\right)+3H_{2}S{O}_4\left(aq\right)\to A{l}_2{(}_S\left(s\right)+3H_2\left(g\right)\\ X0.510mol\\ molA{l}_2{\text{SO}}_4=\text{}\frac{0.510mol{H}_{2\text{}}S{O}_4\text{}\times 1molA{l}_2{\text{SO}}_4}{3mol{H}_2\text{}S{O}_4}\text{}\end{array}$

mol Al₂(SO₄)₃ = 0.170 mol

now we calculate mass Al₂(SO₄)₃ using the following expression

  $\begin{array}{l}\begin{array}{l}massA{l}_2{(S{O}_4)}_3=342.19g/mol\hfill \\ \hfill \\ massA{l}_2{(S{O}_4)}_3=0.170mol\times 342.19g/mol\hfill \\ \hfill \\ massA{l}_2{(S{O}_4)}_3=58.17g\hfill \end{array}\\ \end{array}$