# Stoichiometry Gold III hydroxide

• MichaelXY
In summary: The Attempt at a SolutionIn summary, a chemist makes 20.00g of the KAuCl4 with 25.00 g of the sodium carbonate to prepare a fresh supply of gold III hydroxide. The percent yield if the actual amount of gold III hydroxide produced in the reaction is 10.56g is 80.45%.
MichaelXY

## Homework Statement

This problem is giving me trouble on many levels.
Gold III hydroxide is used for electroplating gold onto other metals. It can be made from the reaction of KAuCl4 (aq) with sodium carbonate and water.

a. Write a balanced equation for this reaction.

b. To prepare a fresh supply of gold III hydroxide, a chemist makes 20.00g of the KAuCl4 with 25.00 g of the sodium carbonate. What is the percent yield if the actual amount of gold III hydroxide produced in the reaction is 10.56g?

## The Attempt at a Solution

Part (a) is my first difficult spot, I was able to come up with

NaCO3 + KAuCl4 + H2O -------> So I figure Au(OH)3 + KCL + CO3, before balancing
3NaCO3 +2KAuCl4+3H20----> 2Au(OH)3 +2KCL+CO2 +6NACl

My first problem was that I do not see how we went from CO3 to CO2, second since CO3 and CO2 are polyatomics, how do we balance.

So I then went on to work the problem.
20 g KAuCl4 = .05292 moles
25 g NaCO3 = .301 mole

So since KAuCL4 is the smaller amount do I say that is the limiting factor? Assuming I am correct, I compute:

20 g KAuCl4 (1 mole KAUCl4/377.877g KAuCl4) (2 mole AuOH3/ 2 mole KAuCl4) (247.991 g AuOH3/ 1 mole AuOH3) = 13.125 g

So Actual over theo gives 80.45%
As any of that correct?
Thanks

Ok, I thought my question here was valid. Read the forum rules, I showed my work, yet no response. I am somewhat puzzled why I received no response. Could someone explain where my question became unreasonable?

Thanks

Your question was not unreasonable and you showed good effort. It just may be that no one could offer any help yet.

MichaelXY said:
13NaCO3 +2KAuCl4+3H20----> 2Au(OH)3 +2KCL+CO2 +6NACl

My first problem was that I do not see how we went from CO3 to CO2, second since CO3 and CO2 are polyatomics, how do we balance.

Na2CO3 is the correct formula and the equation is not balanced and CO2 is not a polyatomic -it is a compound

The decomposition of CO32- is likely due to the acid character of Au3+ (high charge but kind of a big ion) --> Au3+(H2O)6 --> H+ + Au(H2O)5(OH)2+

The H+ then would react with the CO32- and decompose to H2O and CO2.

For the stoichiometry, the smaller no. of moles of reactant does not make it the limiting reactant -
- Use each mol value and determine how many mol of product can be produced.
- How can this information be used to determine which one is the limiting reactant?

Thanks for your help. Much appreciated :)

## 1. What is the chemical formula for Stoichiometry Gold III hydroxide?

The chemical formula for Stoichiometry Gold III hydroxide is Au(OH)3.

## 2. What is the molar mass of Stoichiometry Gold III hydroxide?

The molar mass of Stoichiometry Gold III hydroxide is approximately 267.995 g/mol.

## 3. How is Stoichiometry Gold III hydroxide prepared?

Stoichiometry Gold III hydroxide can be prepared by reacting gold(III) chloride with a strong base, such as sodium hydroxide.

## 4. What is the color of Stoichiometry Gold III hydroxide?

Stoichiometry Gold III hydroxide is a dark brown or black solid.

## 5. What are the common uses of Stoichiometry Gold III hydroxide?

Stoichiometry Gold III hydroxide is commonly used in the production of gold nanoparticles, as a catalyst in organic reactions, and as a precursor for other gold compounds.