this is chemistry subject......please help me do my homework??:(?

Question

the equation for the reaction between elemental magnesium and a solution of gold (III) chloride is

Mg + AuCl3 --> MgCl2 + Au

If a piece of magnesium weighing 1.00 g is placed in a solution that contains 3.00 g of gold (III) chloride dissolved in water, which of the reactants is present in excess? What weight of gold is formed ? how much of the excess reactant is left when the reaction is over?

Answer 1

The balanced chem. equation is as follows:

3Mg + 2AuCl3 --> 3MgCl2 + 2Au

The relative atomic mass of Mg equals Ar = 24.3 g/mol and the relative molar mass of AuCl3 equals Mr = 303.5 g/mol. Now, we can find the moles of each substance:

n = m/Ar, n = 1/24.3 = 0.041 mol Mg, and

n = m/Mr, n = 3/303.5 = 0.01 mol AuCl3 (approx)

3 mol of Mg require 2 mol of AuCl3
0.041 mol of Mg require 2*0.041/3 = 0.027 mol of AuCl3

But we have less (0.01 < 0.027) quantity of AuCl3, so Mg is in excess and AuCl3 reacts fully (the limiting reagent is AuCl3).

2 mol of AuCl3 produce 2 mol of Au, so
0.01 mol of AuCl3 produce 0.01 mol of Au

The relative atomic mass of Au is Ar = 197 g/mol, so the mass of the Au produced is:

m = n*Ar = 0.01*197 = 19.7 g (approx) (19.5 g more exact).

2 mol of AuCl3 react with 3 mol of Mg
0.01 mol of AuCl3 react with 3*0.01/2 = 0.015 mol of Mg

So, when the reaction is over 0.041 - 0.015 = 0.026 mol of Mg is left, with a mass of:

m = n*Ar = 0.026*24.3 = 0.632 g

Answer 2

1. Balance the equation.
2. Find the moles from the masses given.
3. compare their mole ratio to the mole ratio in the reaction. Find out who is in excess.
4. Finish by using the reaction to find the yields of the other components.