Chapter 11, Problem 71A

Interpretation Introduction

Interpretation:

The mass of gold that can be extracted from 25.0 g of sodium cyanide should be calculated.

Concept introduction:

The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

The ratio of amount of any two compounds involved in a chemical reaction in terms of mole is said to be mole ratio, and used as a conversion factor between reactants and products.

Answer to Problem 71A

The mass of gold extracted from 25.0 g of sodium cyanide is $\text{50}\text{.23 g}$.

Explanation of Solution

Given:

The balanced chemical equation:

${\text{4Au(s) + 8NaCN(s) + O}}_2{\text{(g) + 2H}}_{\text{2}}\text{O(l) }\to {\text{ 4NaAu(CN)}}_2\text{(aq) + 4NaOH(aq)}$

The molar mass of sodium cyanide is 49.00 g/mol. The number of moles of sodium cyanide is calculated as shown:

$\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}=\frac{25.0\text{ g}}{\text{49}\text{.0 g/mol}}\\ \text{Number of moles}=0.51\text{ mol}\end{array}$

From the balanced chemical reaction, the mole ratio of gold and sodium cyanide is 4:8 that is 1:2 which means 1 mole of gold reacts with 2 moles of sodium cyanide. So, the number of moles of gold reacted with 0.51 mol of sodium cyanide is:

$0.51\text{ mol of NaCN×}\frac{\text{1 mol of Au}}{\text{2 mol of NaCN}}\text{ = }0.255\text{ mol of Au}$

Now, the mass of gold reacted with 25.0 g of sodium cyanide is calculated as:

The molar mass of gold is 196.97 g/mol. So,

$\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ 0.255\text{ mol}=\frac{\text{Mass}}{\text{196}\text{.97 g/mol}}\\ \text{Mass = 0}\text{.255 mol}\times 196.97\text{ g/mol}\\ \text{Mass = 50}\text{.23 g}\end{array}$

Hence, the mass of gold extracted from 25.0 g of sodium cyanide is $\text{50}\text{.23 g}$.

Interpretation Introduction

Interpretation:

The percentage of the gold in ore should be calculated if the mass of the ore is 150.0 g.

Concept introduction:

The percentage of gold in ore is calculated by dividing the mass of gold extracted by the mass of ore. The formula used is:

$\frac{\text{Mass of gold extracted with NaCN}}{\text{Mass of ore}}\times 100\text{ \%}$

Answer to Problem 71A

The percentage of the gold in ore is $\text{33}\text{.5 \%}$.

Explanation of Solution

The mass of gold extracted from 25.0 g of sodium cyanide is $\text{50}\text{.23 g}$ and the mass of ore is 150.0 g so, the percentage of the gold in ore is calculated as:

$\frac{\text{Mass of gold extracted with NaCN}}{\text{Mass of ore}}\times 100\text{ \%}$

Substituting the values:

$\frac{\text{50}\text{.23 g}}{\text{150 g}}\times 100\text{ \% = 33}\text{.5 \%}$

Hence, the percentage of the gold in ore is $\text{33}\text{.5 \%}$.