Question

6. Write the ICE chart for the reaction of 32.0 g of sulfur and 71.0 g of chlorine: S8 + 4 Cl24S2Cl2 After completing the chart give a. the mass of product b. the mass of agent in excess c. the mass produced if there is an 87.% yield

Answer 1

Answer:

ICE Table Figure

a. 67.37 g $S_2Cl_2$

b. 35.62 g $Cl_2$

c. 58.61 g$S_2Cl_2$

Explanation:

For the ICE table we have to keep in mind that we have 4 moles of $Cl_2$ and 1 mol of $S_8$ and the reactives are consumed, so for $Cl_2$ we will have -4X and for $S_8$ we will have -X. Follow the same logic we will have -4X for $S_2Cl_2$.

a. Mass of the product

Molar mass of $S_8$= 256.52 g/mol

Molar mass of $Cl_2$=70.9 g/mol

Molar mass of $S_2Cl_2$=135.03 g/mol

We have to find the limiting reagent in the reaction:

$S_8+4Cl_2->4S_2Cl_2$

$\frac{32}{256.52}=0.124molS_8$

$\frac{71}{70.9}=1 molCl_2$

Divide by the coefficients in the balanced reaction:

$\frac{0.124}{1}=0.124mol$

$\frac{1}{4}=0.25 mol$

The limiting reagent would be $S_8$

Now is posible to calculate the amount of $S_2Cl_2$ produced:

$0.124molS_8\frac{4molS_2Cl_2}{1 molS_8}\frac{135.03gS_2Cl_2}{1 molS_2Cl_2}=67.37gS_2Cl_2$

b. Mass  in excess

$0.124molS_8\frac{4molCl_2}{1 molS_8}\frac{70.9gCl_2}{1 molCl_2}=35.38gCl_2$

$Excess\hspace{0.1cm}=\hspace{0.1cm}71gCl_2-35.38gCl_2=35.62 gCl_2$

C. 87%Yield

$67.37gS_2Cl_2\frac{87}{100}=58.61gS_2Cl_2$