Textbook Problem

393 views

Gold can be dissolved from gold-bearing rock by treating the rock with sodium cyanide in the presence of oxygen.

4 Au(s) + 8 NaCN(aq) + O₂(g) + 2 H₂O(ℓ) → 4 NaAu(CN)₂(aq) + 4 NaOH(aq)

1. (a) Name the oxidizing and reducing agents in this reaction. What has been oxidized, and what has been reduced?
2. (b) If you have exactly one metric ton (1 metric ton = 1000 kg) of gold-bearing rock, what volume of 0.075 M NaCN, in liters, do you need to extract the gold if the rock is 0.019% gold?

Interpretation Introduction

Interpretation:

The oxidizing and reducing agents in the reaction and the compound /element that has been oxidized and reduced in the given reaction have to be determined.

Concept introduction:

• Oxidizing agent is the reduced (gains electrons) atom to cause another atom to be oxidized.
• Reducing agent is the oxidized (loses electron) atom to cause another atom to be reduced.

Explanation of Solution

Balanced chemical equation for the given reaction is,

  $4A{u}_{(s)}+8NaC{n}_{(aq)}+O_{2(g)}+2H_2{O}_{(l)}\underset{}{\to }4NaAu{(CN}_{}$

Interpretation Introduction

Interpretation:

The volume of $0.075MNaCN$ in liters to extract the gold in according with the given conditions has to be determined.

Concept Introduction:

• The relation between the number of moles and mass of the substance is ,

     $Numberofmole=\frac{Massingram}{Molarmass}$

  $Massingramofthesubs\tan ce=Numberofmole\times Molarmass$

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• $\text{Amount}\text{of substance}\text{=}\text{Concentration}\text{of}\text{substance}\text{×volume}\text{of}\text{the}\text{substance}$
• $Volumeofthesubs\tan ce=\frac{Amountofsubs\tan ce}{Concentrationofthesubs\tan ce}$
• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.