Calculate the total mass of gold needed to react with 4200g of chlorine to make 7112g of gold(iii) chloride?
This is a stoichiometry problem. Here is a worked example I've posted. Just follow the steps. Post your work if you get stuck.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the total mass of gold needed to react with chlorine, we need to determine the stoichiometry of the reaction. The balanced chemical equation for the reaction between gold and chlorine to form gold(III) chloride is as follows:
2Au + 3Cl2 -> 2AuCl3
From the equation, we can see that 2 moles of gold react with 3 moles of chlorine to produce 2 moles of gold(III) chloride.
First, we need to determine the number of moles of chlorine used. To do this, we divide the mass of chlorine by its molar mass:
Number of moles of Cl2 = mass of Cl2 (g) / molar mass of Cl2 (g/mol)
The molar mass of chlorine (Cl2) is calculated by adding the atomic masses of chlorine, which is 35.453 g/mol.
Number of moles of Cl2 = 4200 g / 35.453 g/mol ≈ 118.54 mol
According to the stoichiometry of the reaction, 2 moles of gold react with 3 moles of chlorine. Therefore, to calculate the moles of gold required, we can use the following ratio:
(2 moles of Au / 3 moles of Cl2) = (x moles of Au / 118.54 moles of Cl2)
Simplifying the ratio, we can solve for x:
x = (2 moles of Au / 3 moles of Cl2) * 118.54 moles of Cl2
x ≈ 79.026 mol
Now that we have the number of moles of gold required, we can calculate the mass of gold by multiplying the number of moles by the molar mass of gold (Au):
Mass of Au = number of moles of Au * molar mass of Au
The molar mass of gold is 196.967 g/mol.
Mass of Au = 79.026 mol * 196.967 g/mol ≈ 15,554.25 g
Therefore, the total mass of gold needed to react with 4200 g of chlorine to produce 7112 g of gold(III) chloride is approximately 15,554.25 g.