Tantalum metal is produced by reaction of potassium heptafluorotantalate (K2TaF7) with elemental sodium (Na) in a reactor heated to 850oC. The by-products are potassium fluoride (KF) and sodium fluoride (NaF). The reactor is charged with 244.0 kg of K2TaF7 and 13.0 kg of Na. 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 K2TaF7 remaining:
Enter the number of kg of Na remaining:
Enter the number of kg of Ta present:
Enter the number of kg of KF present:
Enter the number of kg of NaF present:
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just got Ta=42.4. Naf 47.34. nw m confused
To determine the amounts of each compound present in the reactor at the completion of the reaction, we need to calculate the stoichiometry of the reaction and use the given masses of reactants.
The balanced chemical equation for the reaction is:
2 K2TaF7 + 14 Na → 7 NaF + 4 KF + 2 Ta
From the equation, we can see that for every 2 moles of K2TaF7, 14 moles of Na are required to completely react.
1. Calculate the number of moles of K2TaF7:
molar mass of K2TaF7 = (2 * atomic mass of K) + atomic mass of Ta + (7 * atomic mass of F)
= (2 * 39.10 g/mol) + 180.95 g/mol + (7 * 19.00 g/mol)
= 138.20 g/mol
moles of K2TaF7 = mass of K2TaF7 / molar mass of K2TaF7
= 244.0 kg * (1000 g/kg) / 138.20 g/mol
= 1765.30 mol
2. Calculate the number of moles of Na:
molar mass of Na = 22.99 g/mol
moles of Na = mass of Na / molar mass of Na
= 13.0 kg * (1000 g/kg) / 22.99 g/mol
= 565.93 mol
3. Determine the limiting reagent:
To identify the limiting reagent, we compare the mole ratio of Na to K2TaF7. From the balanced equation, 14 moles of Na are required for every 2 moles of K2TaF7.
moles of Na / moles of K2TaF7 = 565.93 mol / 1765.30 mol
= 0.320
Since the mole ratio is less than 7 (which is the ratio of NaF to K2TaF7), Na is the limiting reagent.
4. Determine the amount of each compound formed:
Using the mole ratio from the balanced equation, we can calculate the moles of each product formed.
moles of NaF = 7 * (moles of Na / 2)
= 7 * (565.93 mol / 2)
= 1981.82 mol
moles of KF = 4 * (moles of Na / 2)
= 4 * (565.93 mol / 2)
= 1131.86 mol
moles of Ta = 2 * (moles of Na / 2)
= 2 * (565.93 mol / 2)
= 565.93 mol
5. Convert moles to masses:
mass of NaF = moles of NaF * molar mass of NaF
= 1981.82 mol * (40.00 g/mol)
= 79272.80 g
= 79.27 kg
mass of KF = moles of KF * molar mass of KF
= 1131.86 mol * (58.10 g/mol)
= 65785.61 g
= 65.79 kg
mass of Ta = moles of Ta * molar mass of Ta
= 565.93 mol * (180.95 g/mol)
= 102457.35 g
= 102.46 kg
6. Calculate the remaining mass of K2TaF7 and Na:
mass of K2TaF7 remaining = mass of K2TaF7 initial - mass of Ta formed
= 244.0 kg - 102.46 kg
= 141.54 kg
mass of Na remaining = mass of Na initial - mass of Na used (limiting reagent)
= 13.0 kg - 11.31 kg
= 1.69 kg
Therefore, the reactor contents at the completion of this reaction are:
- The remaining mass of K2TaF7 is 141.54 kg
- The remaining mass of Na is 1.69 kg
- The mass of Ta present is 102.46 kg
- The mass of KF present is 65.79 kg
- The mass of NaF present is 79.27 kg