#10g# of carbon disulfide is combusted with #15.5L# of oxygen gas, which is the reagent in excess?

Answer 1

We have #"dioxygen gas"# in excess by approx. #"17 equiv"#.

I. A stoichiometric equation is required.

#CS_2(l) + 3O_2(g) rarr CO_2(g) + 2SO_2(g)#

And (ii), equivalent quantities of #CS_2(l)#, and #"dioxygen"#.

#"Moles of"# #CS_2=(10.0*g)/(76.14*g*mol^-1)=0.0131*mol#.

Now (depending on your syllabus), #1*mol# of #"Ideal Gas"# occupies #22.7*L# at #"STP"#. If we (REASONABLY) assume ideality, then we have #(15.5*L)/(22.7*L*mol^-1)=0.683*mol*"dioxygen gas"#. And clearly, we have a stoichiometric EXCESS of dioxygen gas.

And thus #"EXCESS"# #O_2=(0.683-3xx0.0131)*mol#

#-=0.644*mol#.

Note that #CS_2# is (i) VERY FLAMMABLE and volatile; and (ii) PEN AND INKS very badly.

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Answer 2

Axel Alvarado

Oxygen gas is in excess.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

#10*g# of carbon disulfide is combusted with #15.5*L# of oxygen gas, which is the reagent in excess?

#10g# of carbon disulfide is combusted with #15.5L# of oxygen gas, which is the reagent in excess?