Question

Gold reacts with chlorine at ${150}^o$C as per balanced chemical equation, $2Au+3C{l}_2\to 2AuC{l}_3$. $10$g of each gold and chlorine are sealed in a container and heated at ${150}^o$C till the reaction is complete. Name the limiting and excess reactants. Also calculate the amount of $AuC{l}_3$ formed and mass of excess reactant left behind. (Atomic masses Au$=196.97$u and Cl$=35.5$u).

Answer 1

$2Au+3C{l}_2\to 2AuC{l}_3$

10gram of each gold and chlorine sealed, let's get the moles of them

moles of $Au=\frac{10}{196.97}=.005$

moles of $C{l}_2=\frac{10}{71}=.140$

To get the limiting reagent divide the moles with respective stochiometry coefficient,after this the element which has less value will be limiting reagent

cofficient of $Au=2\to \frac{0.005}{2}=2.5\times {10}^{-3}$

cofficient of $C{l}_2=3\to \frac{0.140}{3}=46\times {10}^{-3}$

Au will be limiting reagent

At equilibrium-

$2Au+3C{l}_2\to 2AuC{l}_3$

$0.005-2x.140-3x$

Au is limiting reagent so, $2x=.005orx=2.5\times {10}^{-3}$

Put the value of x for $C{l}_2$

Excess rectent left behind $=0.140-3\times 2.5\times {10}^{-3}=0.13925mol$