Answer to Question #107094 in General Chemistry for ABIDAKUN TOMIDE

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: $0.01508 moles Cl_2$ , $1.069 g Cl_2$