If 0.433 mol of gaseous CS2 and 1.62 mol of gaseous H2 are reacted according to the balanced equation at 454 K and in a 40.0 L vessel, what partial pressure (atm) of gaseous CH4 is produced?

Question

If 0.433 mol of gaseous CS2 and 1.62 mol of gaseous H2  are reacted according to the balanced equation at 454 K and in a 40.0 L vessel, what partial pressure (atm) of gaseous CH4  is produced?

CS2(g) + 4H2(g) -> CH4(g) + 2H2S(g)

Molar Mass (g/mol)
CS2 	76.143
H2 	2.016
CH4 	16.043

bobpursley · Answer

figure the moles of the reactants first. THen determine the limiting reageant (CS2). THen determine the moles of methane (.443) produced.

partial pressure=molesmethane*RT/40

Remember that there is still H2 left over, and it will have a partial pressure also, as will the H2S produced, all add to total pressure if you have to calculate that.

Luc · Answer

I tried the method that you provided above and the grading system told me that the answer was wrong. The correct answer is (0.3775). Do you know how to come up with that answer?