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Urea CO(NH2)2 as a solid and water as a gas can be made from the reaction of gaseous carbon dioxide and ammonia. If 12.0g of carbon dioxide reacts with 15.0g of ammonia in a reaction vessel, what is the mole fraction of water vapor in the reaction vessel?

Respuesta :

Hagrid
Urea (CO(NH2)2) synthesis:

CO2 + 2NH3 ===> CO(NH2)2 + H2O

Given:

12.0 grams CO2
15.0 grams of NH3

After the reaction:

moles of CO2 = 12 grams / 44g/mol = 0.27 moles = 0.27/1 = 0.27
moles of NH3 = 15 grams/ 17g/mol = 0.88 moles = 0.88/2 = 0.44

The limiting reactant is CO2. Thus, we will base our calculations on the amount of CO2 available.

moles H2O produced = 0.27 moles CO2 * 1 mol/ 1 mol = 0.27 moles H2O
moles Urea produced = 0.27 moles CO2 * 1 mol/1 mol = 0.27 moles Urea


Total moles in the vessel = excess ammonia + H2O + Urea
                                           = (0.44 - 0.27) + 0.27 + 0.27
                                           = 0.71 moles

The mole fraction of water vapor in the reaction vessel is:

0.27 moles water vapor / 0.71 moles = 0.38 

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