Chapter 18, Problem 18.104QA

Interpretation Introduction

To:

a. Balance the chemical equation.

b. Find out the composition of gehlenite.

Answer to Problem 18.104QA

Solution:

a. The balanced chemical equation is

$Ca{Al}_2{Si}_2{O}_8+{ Ca}_2{Al}_{2}Si{O}_7+3 CaSi{O}_3\to 2 {Ca}_3{Al}_2(Si{O}_4{)}_3$

b. The composition of gehlenite is:

$CaO$	$40.9\%$
${Al}_2{O}_3$	$37.1\%$
$Si{O}_2$	$21.9\%$
$Ca$	$29.2\%$
$Al$	$19.7\%$
$Si$	$10.2\%$

Explanation of Solution

1) Concept:

The reaction is balanced by taking an inventory of the atoms on both sides of the reaction. The  $Ca, Al and Si$ are balanced first by adding appropriate coefficients. The $O$ is balanced last.

The composition of the metals and the metalloid oxides is found out by first finding how many number of moles of the respective metals and oxides are present. The number of moles will give us the amount of the metal or oxide is present, which can be then divided by the total weight to get the percentage composition. The mineral gehlenite can be written in the form of oxides as

${Ca}_2{Al}_{2}Si{O}_7 \to 2 CaO+ {Al}_2{O}_3+Si{O}_2$

Hence, 1 mole of gehlenite contains two moles of calcium oxide, and one mole each of aluminum oxide and silicon oxide. In terms of metals, gehlenite contains two moles of calcium, two moles of aluminum and one mole of silicon.

2) Formula:

$percentage composition= \frac{mass }{total mass} \times 100$

3) Given:

i) Unbalanced chemical reaction is

$Ca{Al}_2{Si}_2{O}_8+{Ca}_2{Al}_{2}Si{O}_7+CaSi{O}_3\to {Ca}_3{Al}_2(Si{O}_4{)}_3$

ii) Molar mass of $Ca=40.078\frac{g}{mol}$

iii) Molar mass of $Al=26.98 \frac{g}{mol}$

iv) Molar mass of $Si=28.08 \frac{g}{mol}$

v) Molar mass of $O=15.999 \frac{g}{mol}$

4) Calculations:

a)

The unbalanced equation is

$Ca{Al}_2{Si}_2{O}_8+{Ca}_2{Al}_{2}Si{O}_7+CaSi{O}_3\to {Ca}_3{Al}_2(Si{O}_4{)}_3$

Taking an inventory of the atoms, we get

$4 Ca+4 Al+4 Si+18 O \to 3 Ca+2 Al+3 Si+12 O$

To balance the $Al$, we add a coefficient 2 in front of ${Ca}_3{Al}_2(Si{O}_4{)}_3$

$Ca{Al}_2{Si}_2{O}_8+{ Ca}_2{Al}_{2}Si{O}_7+CaSi{O}_3\to 2 {Ca}_3{Al}_2(Si{O}_4{)}_3$

Taking an inventory of the atoms, we get

$4 Ca+4 Al+4 Si+18 O \to 6 Ca+4 Al+6 Si+24 O$

To balance the $Ca and Si$, we add a coefficient of 3 in front of $CaSi{O}_3$

$Ca{Al}_2{Si}_2{O}_8+{ Ca}_2{Al}_{2}Si{O}_7+3 CaSi{O}_3\to 2 {Ca}_3{Al}_2(Si{O}_4{)}_3$

Taking an inventory of the atoms, we get

$6 Ca+4 Al+6 Si+24 O \to 6 Ca+4 Al+6 Si+24 O$

The reaction is now completely balanced.

The final balanced reaction is

$Ca{Al}_2{Si}_2{O}_8+{ Ca}_2{Al}_{2}Si{O}_7+3 CaSi{O}_3\to 2 {Ca}_3{Al}_2(Si{O}_4{)}_3$

Molar mass of gehlenite ${ Ca}_2{Al}_{2}Si{O}_7$

$=\left(2 mol Ca\times \frac{40.07g}{1 mol Ca}\right)+2 mol Al\times \left(\frac{26.98g}{1 mol Al}\right)+1 mol Si \times \left(\frac{28.08g}{1 mol Si}\right)+\left(7 mol O \times \frac{15.99g}{1 mol O }\right)$

$molar mass of {Ca}_2{Al}_{2}Si{O}_7=274.11\frac{g}{mol}$

Molar mass of $CaO$

$=\left(1 mol Ca\times \frac{40.07g}{1 mol Ca}\right)+1 mol O\times \left(\frac{15.99g}{1 mol O}\right)$

$molar mass of CaO=56.06\frac{g}{mol}$

Molar mass of $A{l}_2{O}_3$

$=\left(2 mol Al\times \frac{26.98g}{1 mol Al}\right)+3 mol O\times \left(\frac{15.99g}{1 mol O}\right)$

$molar mass of {Al}_2{O}_3=101.93\frac{g}{mol}$

Molar mass of $Si{O}_2$

$=\left(1 mol Si\times \frac{28.08g}{1 mol Si}\right)+2 mol O\times \left(\frac{15.99g}{1 mol O}\right)$

$\therefore molar mass of Si{O}_2=60.06\frac{g}{mol}$

The percentage of $CaO$ in gehlenite is

$\% CaO= \frac{2\times mass of CaO}{molar mass of {Ca}_2{Al}_{2}Si{O}_7 } \times 100= \frac{2\times 56.06 g}{274.11 g} \times 100=40.903 \%$

The percentage of ${Al}_2{O}_3$ in gehlenite is

$\% {Al}_2{O}_3= \frac{mass of {Al}_2{O}_3}{molar mass of {Ca}_2{Al}_{2}Si{O}_7 } \times 100= \frac{101.93\frac{g}{mol}}{274.11\frac{g}{mol}} \times 100=37.185 \%$

The percentage of $Si{O}_2$ in gehlenite is

$\% Si{O}_2= \frac{mass of Si{O}_2}{molar mass of {Ca}_2{Al}_{2}Si{O}_7 } \times 100= \frac{60.06\frac{g}{mol}}{274.11\frac{g}{mol}} \times 100=21.910 \%$

Now, the percentage of calcium metal in gehlenite is

$\% Ca= \frac{2\times mass of Ca}{mass of {Ca}_2{Al}_{2}Si{O}_7 } \times 100= \frac{2\times 40.07\frac{g}{mol}}{274.11\frac{g}{mol}} \times 100=29.236 \%$

The percentage of aluminum metal in gehlenite is

$\% Al= \frac{2\times mass of Al}{mass of {Ca}_2{Al}_{2}Si{O}_7 } \times 100= \frac{2\times 26.98\frac{g}{mol}}{274.11\frac{g}{mol}} \times 100=19.685 \%$

The percentage of silicon metal in gehlenite is

$\% Si= \frac{mass of Si}{mass of {Ca}_2{Al}_{2}Si{O}_7 } \times 100= \frac{28.08\frac{g}{mol}}{274.11\frac{g}{mol}} \times 100=10.244 \%$

Thus the percentages of metal and metalloid oxides are

$CaO$	$40.9\%$
${Al}_2{O}_3$	$37.1\%$
$Si{O}_2$	$21.9\%$
$Ca$	$29.2\%$
$Al$	$19.7\%$
$Si$	$10.2\%$

Conclusion:

The percentage composition of the metals and metalloid oxides is determined using the molar mass of the respective metals and oxides.