Chapter 11, Problem 45A

Smelting When tin(IV) oxide is heated with carbon in a process called smelting, the element tin can be extracted.

   ${\text{SnO}}_{\text{2}}\left(\text{s}\right)\text{+2C}\left(\text{s}\right)\to \text{Sn}\left(\text{l}\right)\text{+2CO}\left(\text{g}\right)$

Interpret the chemical equation in terms of particles, moles, and mass.

Interpretation Introduction

Interpretation:

The given reaction in terms of particles, moles and mass needs to be determined.

${\text{SnO}}_{\text{2}}{}_{\text{(s)}}{\text{ + 2 C}}_{\text{(s)}}\stackrel{}{\to }{\text{ Sn}}_{\text{(l)}}{\text{ + 2 CO}}_{\text{(g)}}$

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can be written as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 45A

${\text{1 molecule SnO}}_{\text{2}}{}_{\text{(s)}}{\text{ + 2 molecules C}}_{\text{(s)}}\stackrel{}{\to }{\text{ 1 molecule Sn}}_{\text{(l)}}{\text{ + 2 molecules CO}}_{\text{(g)}}$

${\text{1 particle SnO}}_{\text{2}}{}_{\text{(s)}}{\text{ + 2 particles C}}_{\text{(s)}}\stackrel{}{\to }{\text{ 1 particle Sn}}_{\text{(l)}}{\text{ + 2 particles CO}}_{\text{(g)}}$

$\begin{array}{l}{\text{SnO}}_{\text{2}}{}_{\text{(s)}}{\text{ + 2 C}}_{\text{(s)}}\stackrel{}{\to }{\text{ Sn}}_{\text{(l)}}{\text{ + 2 CO}}_{\text{(g)}}\\ \text{150}\text{.7g 24}\text{.0 g 118}\text{.7 g 56}\text{.0 g}\end{array}$

Explanation of Solution

According to the given balance chemical reaction:

${\text{SnO}}_{\text{2}}{}_{\text{(s)}}{\text{ + 2 C}}_{\text{(s)}}\stackrel{}{\to }{\text{ Sn}}_{\text{(l)}}{\text{ + 2 CO}}_{\text{(g)}}$

Thus the reaction in terms of molecules:

${\text{1 molecule SnO}}_{\text{2}}{}_{\text{(s)}}{\text{ + 2 molecules C}}_{\text{(s)}}\stackrel{}{\to }{\text{ 1 molecule Sn}}_{\text{(l)}}{\text{ + 2 molecules CO}}_{\text{(g)}}$

Thus the reaction in terms of particles:

${\text{1 particle SnO}}_{\text{2}}{}_{\text{(s)}}{\text{ + 2 particles C}}_{\text{(s)}}\stackrel{}{\to }{\text{ 1 particle Sn}}_{\text{(l)}}{\text{ + 2 particles CO}}_{\text{(g)}}$

The reaction in terms of mass:

Molar mass of ${\text{ SnO}}_{\text{2}}{}_{\text{(s)}}$ = 150.7 g/mol

Molar mass C = 12.0 g/mol

Molar mass Sn = 118.7 g/mol

Molar mass of ${\text{CO}}_{\text{(g)}}$ = 28 g/mol

$\begin{array}{l}{\text{SnO}}_{\text{2}}{}_{\text{(s)}}{\text{ + 2 C}}_{\text{(s)}}\stackrel{}{\to }{\text{ Sn}}_{\text{(l)}}{\text{ + 2 CO}}_{\text{(g)}}\\ {\text{1 moles SnO}}_{\text{2}}{}_{\text{(s)}}\text{×}\frac{\text{150}{\text{.7g SnO}}_{\text{2}}{}_{\text{(s)}}}{{\text{1 mole SnO}}_{\text{2}}{}_{\text{(s)}}}\text{=150}\text{.7g}\\ \text{2 moles C×}\frac{\text{12}\text{.0g C }}{\text{1 mole C}}\text{= 24}\text{.0 g}\\ \text{1 moles Sn ×}\frac{\text{118}\text{.7 g Sn }}{\text{1 mole Sn}}\text{= 118}\text{.7 g}\\ \text{2 moles CO ×}\frac{\text{28}\text{.0 g CO }}{\text{1 mole CO}}\text{= 56}\text{.0 g}\end{array}$

Conclusion

Thus,

$\begin{array}{l}{\text{SnO}}_{\text{2}}{}_{\text{(s)}}{\text{ + 2 C}}_{\text{(s)}}\stackrel{}{\to }{\text{ Sn}}_{\text{(l)}}{\text{ + 2 CO}}_{\text{(g)}}\\ \text{150}\text{.7g 24}\text{.0 g 118}\text{.7 g 56}\text{.0 g}\end{array}$