Gas Stiochiometry?

Question

Urea (H2NCONH2) is used extensively as a nitrogen source in fertilizers.

2NH3 (g) + CO2 (g) --> H2NCONH2 (s) + H2O (g)

Ammonia gas at 223 C and 90. atm flows into a reactor at a rate of 500 L/min. Carbon dioxide at 223 C and 45 atm flows into the reactor at a rate of 600L/min. What mass of urea is produced per minute by this reaction assuming 100% yield?

Answer 1

I keep answering this same question, so you must all go to the same school. Well, ignore the time per minute, as it is equal. First find mols, PV=nRT

(90atm)(500L) = n(0.08206Latm/molK)(496.15K)

45000 = 40.714n

1105.27 = n ( mols NH3 )

(45atm)(600L) = n(0.08206Latm/molK)(496.15K)

27000 = 40.714n

663.16 = n ( mols CO2 )

Now, I know NH3 limits the reaction. Here's how.

1105.27mol NH3 X1mol CO2/2mol NH3 = 552.635 mol CO2

You have more CO2 than that, so NH3 drives the reaction.

1105.27 mol NH3 X 1mol Urea/2mol NH3 X 60.062g Urea/1mol Urea = 33192.36 grams Urea produced per minute.

Answer 2

First you need to know the moles of the reactants.

Use the ideal gas equation PV=nRT.

You have the pressure, volume, (assume 1 minute) temperature. Solve for n = moles.

Donn't forget to convert C into K. Use 0.082 for the R.

Now that you have the moles of ammonia and CO2, you can use the chemical equation that you have.
The moles of urea need to be calculated both fo rammonia and CO2 to fine the limiting reactant. Once you have that, convert the moles of urea to mass.

Answer 3

the 1st ingredient to do right it rather is to jot down a balanced eqation for the reaction: Xe + 2F2 ---> XeF4 next calculate the form of Mol of Xe and F2 you have interior the equipment. to do this use the suited gasoline regulation PV=nRT. you be responsive to V=20 l, T=4 hundred C (673 ok), and the pressures for each Xe and F2. because of the fact the pressures are an analogous, the ensuing value often is the form of mol for the two Xe and F2. (0.5 atm)*(20 l) = n*(0.08206 mol*ok/l*atm)*(673K) n=0.181 mol of Xe and 0.181 mol F2 because of the fact two times as lots F2 would be ate up interior the reaction as Xe, the F2 often is the proscribing reagent right here and the form of mol of product produced would be = 0.181/2 = 0.0905 mol XeF4. Now multiply this by the MW of XeF4 to get the mass of XeF4 if the reaction gave one hundred% yeild: 0.0905 mol XeF4 x (207.3 g/mol) = 18.seventy six g

Answer 4

woo ya that's fun I forget how to do this but I know it has to do with molarities