Question

Uranium is distributed in a form called yellow cake, which is made from

uranium ore. In the second step of the reactions that form yellow cake from uranium ore, uranyl sulfate, UO2SO4, is converted to (NH4)2U2O7.

2UO2SO4 + 6NH3 + 3H2O → (NH4)2U2O7 + 2(NH4)2SO4

a. What is the maximum mass, in kilograms, of (NH4)2U2O7 that can be formed from the reaction of 100 kg of water and 100 kg of ammonia with 481 kg of UO2SO4?

b. Explain why two of these substances are in excess and one is limiting

Answer 1

2UO2SO4 + 6NH3 + 3H2O → (NH4)2U2O7 + 2(NH4)2SO4

from the equation,

2mol UO2SO4 = 6 mol NH3 = 3 mol H2O

NO of mol of NH3 = 100*10^3/17 = 5882.35 mol

No of mol of water = 100*10^3/18 = 5555.6 mol

No of mol of UO2SO4 = 481*10^3/366.09 = 1314 mol

from the stoichiometric coefficients.

SO that limiting reagent is UO2SO4

(the reagent which is completely used up in the reaction called limiting reagent)

No of mol of (NH4)2U2O7 produced in the reaction

= 1314/2 = 657 mol

mass of (NH4)2U2O7 produced = 657*624.1305 = 410.06 kg