Methane and hydrogen sulfide form when hydrogen reacts with carbon disulfide. identify the excess reagent and calculate how much remains after 36 l of h2 reacts with 12 l of cs2. 4h2(g) + cs2(g) → ch4(g) + 2h2s(g)

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

2b2t

explanation:

2b2t