What volume of ammonia gas at 25 C and 1.5 atm is needed to make 500g of urea?

Question

Urea, NH2CONH2, is manufactured from ammonia and carbon dioxide according to the following equation:

2 NH3 (g) + CO2 (g)  NH2CONH2 (aq) + H2O (l)

What volume of ammonia gas at 25o C and 1.5 atm is needed to make 500g of urea?

Data: Atomic weights N = 14.01, O = 16.00, H = 1.01, C = 12.01

PV = nRT where R = .0821 atm.L / K.mol

Molarity, mol/L = 1000 n / V where n = moles and V = volume in mL

Please help me!!!

Answer 1

first you need to figure out how many mole of ammonia you'll need:

500g urea (MW=60 g/mole) = 8.33 moles

2NH3 --> CON2H4

so you'll need 16.66 moles NH3

The V = nRT/P = 16.66moles(0.821atmL/K•mole)
(25º + 273K)/1.5 atm = 271.73 L

Answer 2

First, find the atomic weight of urea by adding the weight of all the nitrogens, oxygens and carbons in the urea together. Then use this to determine how many moles of nitrogen you need to make 500 grams of urea. Use stoichiometry. 2 moles of NH3=1 mole of urea.

Finally, after you've determined the number of moles of NH3 you need, you have three variables and can solve for volume using the ideal gas law (PV=nRT). Use the equation V=nRT/P.

Answer 3

First, convert the mass of urea into moles using its molar mass.

From the balanced equation, you see that you need twice as many moles of ammonia as you make of urea, so multiply the moles of urea by 2 to get the moles of NH3 needed.

Next, use the ideal gas law to calculate the volume of that many moles of NH3 under the conditions given.

Answer 4

the molecular mass of urea is
14.01+2.02+12+16+14.01+2.02 =60.06g
500g of Urea correspond to 500/60.06 =8.325 moles of urea

you see from the equation that youneed 2 moles of NH3 for 1 mole of urea

So , you need 8.325*2 =16.65 moles of NH3

Use pV = nRT so V=nRT/p

V = 16.65*0.0821*(273+25)/1.5 =271.6 Liters