Answer:
There will be produced 9L of CH4 and 18 L of H2S. There will remain 3 L of CS2
Explanation:
Step 1: Data given
volume of H2 = 36.00 L
volume of CS2 = 12 L
Step 2 = the balanced equation
4H2(g) + CS2(g) → CH4(g) + 2H2S(g)
Step 3: Calculate number of moles of H2
1 mol = 22.4 L
36 L = 1.607 mol
Step 4: Calculate moles of CS2
1 mol = 22.4 L
12 L = 0.5357 moles
Step 5: Calculate the limiting reactant
For 4 moles of H2 we need 1 mol of CS2 to produce 1 mol of CH4 and 2 moles of H2S
H2 is the limiting reactant. It will completely be consumed. ( 1.607 moles)
CS2 is in excess. There will react 1.607/4 = 0.40175 moles
There will remain 0.5357 - 0.40175 = 0.13395 moles of CS2
0.13395 moles of CS2 = 3 L
Step 6: Calculate products
For 4 moles of H2 we need 1 mol of CS2 to produce 1 mol of CH4 and 2 moles of H2S
For 1.607 moles of H2 we'll have 0.40175 moles of CH4 (= 9L) and 0.8035 moles of H2S =(18L)
There will be produced 9L of CH4 and 18 L of H2S. There will remain 3 L of CS2
