How many liters of carbon dioxide can be produced if 37.8 grams of carbon disulfide react with excess oxygen gas at 28.85 degrees Celsius and 1.02 atmospheres? CS2(l) + 3O2(g) yields CO2(g) + 2SO2(g) 2.78 liters 5.95 liters 11.9 liters 12.2 liters

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How many liters of carbon dioxide can be produced if 37.8 grams of carbon disulfide react with excess oxygen gas at 28.85 degrees Celsius and 1.02 atmospheres? CS2(l) + 3O2(g) yields CO2(g) + 2SO2(g) 2.78 liters 5.95 liters 11.9 liters 12.2 liters

Chemistry
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@jammy987 you know right away from the question what your limiting reagent is can you tell what that is?
@photon336 so the answer would be b

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Carbon di sulfide is our limiting reagent so you need use that to calculate the moles of carbon dioxide produced. do you know how to do this?
\[\frac{ 1 mol CS _{2} }{76 g } * 37.8 g CS _{2} = moles CS _{2}\]
you first need to ensure that your chemical equation is balanced first then calculate the number of moles of CS2 once you get the moles of CS2 you must multiply by the molar ratio to get how much CO2 was produced. \[Mol CS _{2} * \frac{ CO _{2}}{ CS _{2} } = Mol CO _{2}\] then tell me what you get from this.
Then when you get the moles of CO2 use the formula below. \[PV = nRT \] You're solving for volume. You know pressure, Temperature, and R= 0.082 and n. But make sure your temperature is in Kelvin not Celsius to do that its K = C+273 \[\frac{ nRT }{ P }= Volume \in liters \] for carbon dioxide
yes so it would be c?@photon336
I didn't get c, although I got close to that. How did you get c?
@JoannaBlackwelder oh it was d
Yeah, that's what I got. :-)

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