Chapter 6.2, Problem 3RQ

(a)

Interpretation Introduction

Interpretation: The mass of BaCl₂ needs to be determined.

Concept Introduction: The mass can be calculated from number of moles and molar mass as follows:

  $m=n\times M$

Here, n is number of moles and M is molar mass.

Answer to Problem 3RQ

31.24 g

Explanation of Solution

  ${\text{Molar mass of BaCl}}_{\text{2}}\text{=Atomic mass of Ba+2×Atomic mass of Cl}$

  $=137.32+2\times 35.5$

  $=208.32$

The number of moles is related to mass and molar mass as follows:

  ${\text{Number of moles of BaCl}}_{{}_{\text{2}}}\text{=}\frac{\text{mass}}{\text{Molar mass}}$

  $0.15=\frac{\text{mass}}{\text{208}\text{.32}}$

Thus,

  ${\text{mass of BaCl}}_{{}_{\text{2}}}=0.15\text{ mol}\times 208.32\text{ g/mol}$

  $=31.24\text{ g}$

(b)

Interpretation Introduction

Interpretation: The mass of CO₂ needs to be determined.

Concept Introduction: The mass can be calculated from number of moles and molar mass as follows:

  $m=n\times M$

Here, n is number of moles and M is molar mass.

Answer to Problem 3RQ

88 g

Explanation of Solution

  ${\text{Molar mass of CO}}_{\text{2}}\text{=Atomic mass of Carbon+2×Atomic mass of Oxygen}$

  $\begin{array}{l}=12+2\times 16\\ =44\text{ g/mol}\end{array}$

The given mass can be calculated as follows:

  ${\text{Number of moles of CO}}_{{}_{\text{2}}}\text{=}\frac{\text{mass}}{\text{Molar mass}}$

  $2.0=\frac{\text{ mass}}{44}$

  ${\text{ mass of CO}}_{{}_{\text{2}}}\text{=2}\text{.0×44 g}$

  $=88\text{ g}$

(c)

Interpretation Introduction

Interpretation: The mass of HNO₃ needs to be determined.

Concept Introduction: The mass can be calculated from number of moles and molar mass as follows:

  $m=n\times M$

Here, n is number of moles and M is molar mass.

Answer to Problem 3RQ

220.5 g.

Explanation of Solution

  ${\text{Molar mass of HNO}}_{\text{3}}\text{=Atomic mass of Hydrogen+Atomic mass of Nitrogen+3×Atomic mass of Oxygen}$

  $\begin{array}{l}=1+14+3\times 16\\ =63\end{array}$

The number of moles can be calculated as follows:

  ${\text{Number of moles of HNO}}_{{}_{\text{3}}}\text{=}\frac{\text{ mass}}{\text{Molar mass}}$

  $3.50=\frac{\text{ mass}}{63}$

  ${\text{mass of HNO}}_{{}_{\text{3}}}=3.50\times 63$

  $=220.5gm$

(d)

Interpretation Introduction

Interpretation: The mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_4$ needs to be determined.

Concept Introduction: The mass can be calculated from number of moles and molar mass as follows:

  $m=n\times M$

Here, n is number of moles and M is molar mass.

Answer to Problem 3RQ

  $\text{mass of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_4=\text{85}\text{.8 g}$

Explanation of Solution

  $\text{Molarmassof}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{{}_{\text{4}}}\text{=2×(Atomic mass of Nitrogen+4×Atomic mass of Hydrogen)+}$

  $\text{Atomic mass of Sulphur+4×Atomic mass of Oxygen}$

Putting the values,

$\text{Molar mass of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{{}_{\text{4}}}=2\times (14+4\times 1)+32+4\times 16$

  $=2\left(18\right)+96$

  $=36+96$

$=132$

The number of moles is related to mass and molar mass as follows:

  $\text{Number of moles of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_4$

  $=\frac{\text{Given mass}}{\text{Molar mass}}$

  $0.65=\frac{\text{Given mass}}{\text{132}}$

  $\text{Given mass of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{{}_{\text{4}}}=0.65\text{ mol}\times 132\text{ g/mol}$

  $=85.8\text{ g}$