Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is:
S2(g)+C(s)<-> CS2(g) Kc=9.40 at 900K
How many grams of CS2(g) can be prepared by heating 16.8 moles of S2(g) with excess carbon in a 9.80 L reaction vessel held at 900 K until equilibrium is attained?
http://chemwiki.ucdavis.edu/Physical_Chemistry/Chemical_Equilibrium/Le_Chatelier%27s_Principle/Ice_Tables
To determine the grams of CS2(g) that can be prepared, we need to use the given equilibrium constant (Kc) and the stoichiometry of the chemical equation.
1. Start by writing the balanced chemical equation:
S2(g) + C(s) <--> CS2(g)
2. Since the mole ratio of S2 to CS2 is 1:1 based on the balanced equation, the number of moles of CS2 formed will be equal to the number of moles of S2 used.
3. Convert the moles of S2 to grams using its molar mass. The molar mass of S2 is 32.07 g/mol.
moles of S2 = 16.8 mol
grams of S2 = 16.8 mol * 32.07 g/mol
4. Now we need to use the equilibrium constant (Kc) to find the moles of CS2.
Kc = [CS2] / [S2]
Kc = 9.40
Since the initial concentration of S2 is equal to its equilibrium concentration:
[S2] = 16.8 mol / 9.80 L
Substituting the values into the equation:
9.40 = [CS2] / (16.8 mol / 9.80 L)
Solve for [CS2]:
[CS2] = 9.40 * (16.8 mol / 9.80 L)
5. Finally, convert the moles of CS2 obtained from the previous step to grams using its molar mass. The molar mass of CS2 is 76.14 g/mol.
moles of CS2 = [CS2]
grams of CS2 = [CS2] * 76.14 g/mol
Plug in the value of [CS2] you obtained from the previous step into the equation to calculate the grams of CS2(g) that can be prepared.