Question

If 1.250 of AU and 1.744 of CL2 were mixed determine which of the reactants is in excess

2. Calculate the excess amount

Answer 1

$2Au + 3Cl_2 \rightarrow 2AuCl_3$

1) Find moles of Au

$1.250 g Au (\frac{1 mole Au}{196.97 g Au})=0.006346 mol$

2) Find moles of Cl₂

$1.744 g Cl_2 (\frac{1 mol Cl_2}{2\times 35.45 g Cl_2})=0.02460 mol$

3) Compere the values:

$\frac{moles Au}{2}$ and $\frac{moles Cl_2}{3}$

$\frac{0.006346 mol Au}{2}$ and $\frac{0.02460 mol Cl_2}{3}$

$0.003173 mol Au < 0.008200 mol Cl_2$

Au - is a limiting reactant

Cl_2 is in excess.

3) Use moles of Au to calculate the moles of Cl_2 consumed in the reaction

$0.006346 mol Au (\frac{3 mol Cl_2}{2 mol Au}) = 0.009519 mol Cl_2$

Find the moles of Cl_2 left after the reaction (the excess amount):

$moles Cl_2 = 0.02460 -0.009519 = 0.01508 mol Cl_2$

Find the mass of $Cl_2$ left after the reaction:

$0.01508 mol Cl_2 (\frac{2\times 35.45 g Cl_2}{1 mol Cl_2}) =1.069 g Cl_2$

Answer: excess amount: $0.01508 moles Cl_2$ , $1.069 g Cl_2$