Chapter 7, Problem 7.91QA

Interpretation Introduction

To calculate:

a. The amount of $NaCN$ needed to extract gold as $NaAu{\left(CN\right)}_2$ from $1$ metric ton of rock.

b. The amount of zinc which is required to convert $NaAu{\left(CN\right)}_2$ to metallic gold.

c. The volume of gold block in $c{m}^3$

Answer to Problem 7.91QA

Solution:

a. The amount of $NaCN$ needed to extract gold as $NaAu{\left(CN\right)}_2$ from $1$ metric ton of rock is $45 g$

b. The amount of zinc which is required to convert $NaAu{\left(CN\right)}_2$ to metallic gold is $15 g$

c. The volume of gold block in $c{m}^3$ is $4.66 c{m}^3$

Explanation of Solution

1) Concept:

The overall process describes the extraction of gold from the rock using $NaAu{\left(CN\right)}_2$

The process occurs in two steps:

$4Au\left(s\right)+8NaCN\left(aq\right)+O_2\left(g\right)+2H_{2}O\left(l\right)\to 4NaAu(CN{)}_2+4NaOH(aq)$

$2NaAu\left(CN{)}_2\right(aq)+Zn(s)\to 2Au(s)+{Na}_2[Zn\left(CN{)}_4\right]\left(aq\right)$

Both these reactions are balanced.

In a balanced chemical equation, if we know the amount of reactant, the amount of product can be calculated using stoichiometry of the reactions mentioned above.

2) Formula:

i) $\% composition=\frac{mass of an element}{total mass} \times 100$

ii) $d=\frac{m}{V}$

3) Given:

The two steps involved to extract gold from the rock:

i) $4Au\left(s\right)+8NaCN\left(aq\right)+O_2\left(g\right)+2H_{2}O\left(l\right)\to 4NaAu(CN{)}_2+4NaOH(aq)$

ii) $2NaAu\left(CN{)}_2\right(aq)+Zn(s)\to 2Au(s)+{Na}_2[Zn\left(CN{)}_4\right]\left(aq\right)$

iii) The mass of rock = ${1 metric ton=10}^3 kg$

iv) The density of gold = $19.3g/c{m}^3$

v) Mass % for gold = $0.009\%$

4) Calculation:

a. Mass of rock

$m={10}^{3}kg\times \frac{{10}^{3}g}{1kg}={10}^{6}g$

Mass percent of gold $=0.009\%$

$0.009=\frac{mass of gold}{mass of rock}\times 100$

$0.009=\frac{mass of gold}{{10}^{6}g}\times 100$

Mass of gold extracted $=90.0 g$

Number of moles of Au is calculated as:

$90g Au\times \frac{1molAu}{197g Au}=0.457 mol Au$

From the first given equation, the mole ratio of $Au$ and $NaCN$ is $4:8$ which is same as $1:2$

Moles of $NaCN$ reacted is calculated as:

$0.457mol Au\times \frac{2 mol NaCN}{1 mol Au}=0.914 mol NaCN$

Converting mol of $NaCN$ into mass in grams, we get,

$0.914 mol NaCN\times \frac{49g NaCN}{1 mol}=45 g NaCN$

b. The balanced chemical equation (for 1^st step) is:

$4Au\left(s\right)+8NaCN\left(aq\right)+O_2\left(g\right)+2H_{2}O\left(l\right)\to 4NaAu(CN{)}_2+4NaOH(aq)$

Mole ratio of  $NaCN:NaAu(CN{)}_2=8:4=2:1$

So, the number of moles of $NaAu(CN{)}_2$ can be calculated as:

$0.914molNaCN\times \frac{1 mol NaAu(CN{)}_2}{2 molNaCN}=0.457molNaAu(CN{)}_2$

The balanced chemical equation (for 2^nd step) is:

$2NaAu\left(CN{)}_2\right(aq)+Zn(s)\to 2Au(s)+{Na}_2[Zn\left(CN{)}_4\right](aq$)

Mole ratio of  $NaAu{\left(CN\right)}_2:Zn is 1:2$

So, the number of moles of Zn is calculated as:

$0.457mol NaAu(CN{)}_2\times \frac{1mol Zn}{2mol NaAu(CN{)}_2}=0.229 mol Zn$

Converting $0.914 mol Zn$ into mass in grams:

$0.229 mol Zn\times \frac{65.38g Zn}{1 mol Zn}=15g Zn$

c. The mass of gold extracted = $90g$

The density of gold = $19.3g/c{m}^3$

Volume of gold ingot is calculates as:

$V= \frac{m}{d}=\frac{90g}{19.3g/{cm}^3}=4.66 {cm}^3$

Conclusion

All the quantitative information about the products can be obtained from the balanced equation.