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

ICE Table Figure

a. 67.37 g
`S_2Cl_2`

b. 35.62 g
`Cl_2`

c. 58.61 g
`S_2Cl_2`

Step-by-step 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`

`(32)/(256.52)=0.124molS_8`

`(71)/(70.9)=1 molCl_2`

Divide by the coefficients in the balanced reaction:

`(0.124)/(1)=0.124mol`

`(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(4molS_2Cl_2)/(1 molS_8)(135.03gS_2Cl_2)/(1 molS_2Cl_2)=67.37gS_2Cl_2`

b. Mass in excess

`0.124molS_8(4molCl_2)/(1 molS_8)(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(87)/(100)=58.61gS_2Cl_2`