Question

Can someone eplain how to do this ? : identify the limiting reagent and the volume of CO2 formed when 11 L CS2 reacts with 18 L O2 to produce SO2 gas at STP. CS2(g)+3O2(g) → CO2(g)+2SO2(g)

Answer 1

Use stoichiometry to calculate the volume produced using each reactant. The limiting reactant will be the smaller volume produced.

Answer 2

There are 22.4L of ideal gas at STP for each mole.

Answer 3

im not sure how i would do that . can you please answer it and also explain how you found the answer i found the limiting reagent but i don't know how to find the volume of CO2

Answer 4

I don't know how you could figure put the limiting reactant without finding the volume of CO2.

Answer 5

Or at least the moles.

Answer 6

i used the BCA chart .

Answer 7

Ok. Since we are at STP, I would convert by 11 L of CS2 (1 mol CS2/22.4L CS2) (1 mol CO2/1 mol CS2) (22.4 L CO2/1 mol CO2) = 11 L of CO2

Answer 8

Using the same idea, can you calculate the liters of CO2 produced from 18L of O2?

Answer 9

yes i did it and it came out to be 18 L of CO2

Answer 10

It shouldn't be 18L since the mole ratio from the balanced reaction is 3 O2 to 1 CO2.

Answer 11

Ok let me re try with that in mind

Answer 12

Great

Answer 13

:)

Answer 14

the answer i got is 6 L of CO2

Answer 15

Awesome, that is what I get.

Answer 16

Thank you soo much ! really helped out a lot

Answer 17

So, can you tell from this calculation which is limiting?

Answer 18

No worries, glad to help.