Gold reacts with chlorine at 150oC as per balanced chemical equation, 2Au+3Cl2→2AuCl3. 10g of each gold and chlorine are sealed in a container and heated at 150oC till the reaction is complete. Name the limiting and excess reactants. Also calculate the amount of AuCl3 formed and mass of excess reactant left behind. (Atomic masses Au=196.97u and Cl=35.5u).

Gold reacts with chlorine at 150oC as per balanced chemical equation, 2Au+3Cl2→2AuCl3. 10g of each gold and chlorine are sealed in a container and heated at 150oC till the reaction is complete. Name the limiting and excess reactants. Also calculate the amount of AuCl3 formed and mass of excess reactant left behind. (Atomic masses Au=196.97u and Cl=35.5u).

Solution=
2Au+3Cl2→2AuCl3
10gram of each gold and chlorine sealed, let's get the moles of them
moles of Au=10/196.97=.005
moles of Cl2=10/71=.140

To get the limiting reagent divide the moles with respective stochiometry coefficient, after this the element which has less value will be limiting reagent
coefficient of Au=2→0.005/2=2.5×10^−3
coefficient of Cl2=3→0.140/3=46×10^−3

Au will be limiting reagent

At equilibrium-
2Au+3Cl2→2AuCl3
0.005−2x.140−3x
Au is limiting reagent so, 2x=.005 or x=2.5×10−3
Put the value of x for Cl2

Excess reactant left behind =0.140 −3×2.5×10^−3=0.13925 mol