Carbon disulfide is prepared by heating sulfur and charcoal. the chemical equation is s2(g)+c(s)↽−−⇀cs2(=9.40 at 900 k how many grams of cs2(g) can be prepared by heating 12.5 mol s2(g) with excess carbon in a 5.30 l reaction vessel held at 900 k until equilibrium is attained?

The amount of carbon sulfide prepared is 859.1724 grams.

Explanation:

To calculate the concentration of , we use the equation:

     .....(1)

We are given:

Moles of = 12.5 moles

Volume of solution = 5.3 L

Putting values in above equation, we get:

For the reaction of sulfur and carbon, the equation follows:

At       2.36      

At      2.36 - x               x

We are given:

Putting values in above equation, we get:

Molarity of = 2.133 M

Volume of solution = 5.30 L

Putting values in equation 1, we get:

Moles of carbon sulfide = 11.3049 mol

Molar mass of carbon sulfide = 76 g/mol

Putting values in above equation, we get:

Hence, the amount of carbon sulfide prepared is 859.1724 grams.

the answer for sodium is 0.186

the answer for rubidium is 0.251