Uranium metal can be produced by reaction of uranium tetrafluoride (UF4) with calcium metal (Ca). The by-product of this reaction is calcium fluoride (CaF2). The reactor is charged with 292.0 kg of UF4 and 47.0 kg of calcium. Assume that the reaction goes to completion.
The reactor contents at the completion of this reaction will be the following:
Enter the number of kg of UF4 remaining:
- Enter the number of kg of Ca remaining:
- Enter the number of kg of U present:
- Enter the number of kg of CaF2 present:
3 answers
This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants. This is a standard LR problem. What do you not understand?
If is limit what a reaction with reagent and excess but don't know how to start this exercise, please help me
UF4 + 2Ca ==> 2CaF2 + U
mols UF4 initially = grams/molar mass = 292,000/314 = approx 930 but you need to do this more accurately. All of the following calculations are estimates; recalculate those also.
mols Ca = grams/atomic mass = 47,000/40.1 = estimated 1180
Using the coefficients in the balanced equation, convert mols UF4 to mols U. That's 930 x (1 mol U/1 mol UF4) = estimated 930 mols U.
Do the same for converting mols Ca to mols U. That's 1180 x (1 mol U/2 mols Ca) = 1180 x 1/2 = estimated 590 mols U.
You can see that mols U don't agree; therefore, one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that value is the LR. Therefore, Ca is the limiting reagent and estimated 590 mols U will be formed.
g U formed = mols x atomic mass = 590 x 238 = estimated 140,000 grams U.
Enter the number of kg of U present:
kg U = 140 from above. This is estimated only.
Enter the number of kg of Ca remaining:
Since Ca is the LR, all of it will be consumed and none will be remaining.
Enter the number of kg of CaF2 present:
Convert mols Ca to mols CaF2. That's estimated 1180 mols Ca x (2 mols CaF2/2 mols Ca) = estimated 1180 mol CaF2.
kg CaF2 present = 1180 mols x molar mass CaF2 x (1 kg/1000g) = ?
Enter the number of kg of UF4 remaining:
Use the coefficients to convert mols Ca used to mols UF4 used.
1180 mols Ca x (1 mol UF4/2 mol Ca) = estimated 590 mols UF4 used.
mols UF4 initially = 930 mols.
mols UF4 used = 590 mols.
mols UF4 remaining = 930-590 = estimated 340 mols.
kg UF4 remaining = estimated 340 x molar mass UF4 x (1 kg/1000g) = ?
mols UF4 initially = grams/molar mass = 292,000/314 = approx 930 but you need to do this more accurately. All of the following calculations are estimates; recalculate those also.
mols Ca = grams/atomic mass = 47,000/40.1 = estimated 1180
Using the coefficients in the balanced equation, convert mols UF4 to mols U. That's 930 x (1 mol U/1 mol UF4) = estimated 930 mols U.
Do the same for converting mols Ca to mols U. That's 1180 x (1 mol U/2 mols Ca) = 1180 x 1/2 = estimated 590 mols U.
You can see that mols U don't agree; therefore, one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that value is the LR. Therefore, Ca is the limiting reagent and estimated 590 mols U will be formed.
g U formed = mols x atomic mass = 590 x 238 = estimated 140,000 grams U.
Enter the number of kg of U present:
kg U = 140 from above. This is estimated only.
Enter the number of kg of Ca remaining:
Since Ca is the LR, all of it will be consumed and none will be remaining.
Enter the number of kg of CaF2 present:
Convert mols Ca to mols CaF2. That's estimated 1180 mols Ca x (2 mols CaF2/2 mols Ca) = estimated 1180 mol CaF2.
kg CaF2 present = 1180 mols x molar mass CaF2 x (1 kg/1000g) = ?
Enter the number of kg of UF4 remaining:
Use the coefficients to convert mols Ca used to mols UF4 used.
1180 mols Ca x (1 mol UF4/2 mol Ca) = estimated 590 mols UF4 used.
mols UF4 initially = 930 mols.
mols UF4 used = 590 mols.
mols UF4 remaining = 930-590 = estimated 340 mols.
kg UF4 remaining = estimated 340 x molar mass UF4 x (1 kg/1000g) = ?
3 answers