Chapter 15, Problem 15.37P

Interpretation Introduction

Interpretation:

The amount of BaO that can be added to 1 kg of SiO₂ under given conditions should be determined and comparison should be done with addition of Li₂O in SiO₂.

Concept Introduction:

While glass formation it is important to study chemical, structural and other factors that decide the limits within any type of material that can exist. During the manufacturing of glass the parameters mandatory are the stability of glass, physical nature of glass, inorganic elements and compounds, viscosity, etc. even parameters like rate of cooling, crystallization are very important.

Answer to Problem 15.37P

The amount of BaO that is 1257 g is more than $L{i}_{2}O$ that is 245 g for 1000 g of $Si{O}_2$ for making glass.

Explanation of Solution

Given:

Ratio of O:S = 2.5

  $\frac{\text{O}}{\text{Si}}\text{=}\frac{\text{(1}\frac{\text{O ions}}{\text{BaO}}\text{)(fn)+(2}\frac{\text{O ions}}{{\text{SiO}}_{\text{2}}}\text{)(1-fn)}}{\text{(1}\frac{\text{Si ion}}{{\text{SiO}}_{\text{2}}}\text{)(1-fn)}}$

Where,

fn = mole fraction of BaO

  $(1-fn)$ = mole fraction of $Si{O}_2$

2.5 = $\frac{\text{(1}\frac{\text{O ions}}{\text{BaO}}\text{)(fn)+(2}\frac{\text{O ions}}{{\text{SiO}}_{\text{2}}}\text{)(1-fn)}}{\text{(1}\frac{\text{Si ion}}{{\text{SiO}}_{\text{2}}}\text{)(1-fn)}}$

  $fn=0.33$

The molecular weight of

Silica = 60.08 g/mol

BaO = 137.3 g/mol+16g/mol

=153.3 g/mol

Glass to be produce with

BaO <_ 33.0 mol%

Weight of % BaO is

Weight % BaO =

  $\begin{array}{l}=\frac{(fn)(153.3)}{(fn)(153.3)+(1-fn)(60.08)}\times 100\\ =\frac{(0.33)(153.3)}{(0.33)(153.3)+(1-0.33)(60.08)}\\ =55.69\%\end{array}$

Amount of BaO required = x grams in 1000 g of $Si{O}_2$.

Weight of BaO =

  $\begin{array}{l}=\frac{x}{x+1000}\times 100\\ \therefore \frac{55.69}{100}=\frac{x}{x+1000}\\ \therefore x=1257g\end{array}$

Now, for mole fraction of $L{i}_{2}O$

  $\frac{O}{Si}=\frac{(1\frac{O_{ions}}{L{i}_{2}O})(fn)+(2\frac{O_{ions}}{Si{O}_2})(1-fn)}{(1\frac{Siion}{Si{O}_2})(1-fn)}$

2.5= $\frac{O}{Si}=\frac{(1\frac{O_{ions}}{L{i}_{2}O})(fn)+(2\frac{O_{ions}}{Si{O}_2})(1-fn)}{(1\frac{Siion}{Si{O}_2})(1-fn)}$

  $f_L=0.33$

Molecular weight of silica = 60.08 g/mol

And of $L{i}_{2}O=(2\times 6.94+16)$ g/mol

= 29.88 g/mol

Adding $L{i}_{2}O$<_33.0 mol%

  $wt.\%L{i}_{2}O=\frac{(f_L)(29.88)}{(f_L)(29.88)+(1-f_L)(60.08)}\times 100$

  $\begin{array}{l}=\frac{(0.33)(29.88)}{(0.33)(29.88)+(1-0.33)(60.08)}\\ =19.7\%\end{array}$

Let the amount of $L{i}_{2}O$ require = x g in 1000 g of $Si{O}_2$.

Amount is

Weight % of $L{i}_{2}O$ =

  $\begin{array}{l}=\frac{x}{x+1000}\times 100\\ \frac{19.7}{100}=\frac{x}{x+1000}\\ x=245g\end{array}$

Conclusion

Thus, the amount of BaO that is 1257 g is more than $L{i}_{2}O$ that is 245 g for 1000 g of $Si{O}_2$ for making glass.