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 208.0 kg of K2TaF7 and 28.0 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:
a)the number of kg of Na remaining:
b)the number of kg of Ta present:
c)the number of kg of KF present:
d)the number of kg of NaF present:
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To find the number of kilograms of each substance present at the completion of the reaction, we need to use stoichiometry.
First, let's write out the balanced chemical equation for the reaction:
2K2TaF7 + 14Na → 7Ta + 14NaF + 4KF
From the balanced chemical equation, we can see that for every 2 moles of K2TaF7, we get 7 moles of Ta, 14 moles of NaF, and 4 moles of KF.
Now, let's convert the masses of K2TaF7 and Na to moles:
Molar mass of K2TaF7:
(2 * atomic mass of K) + atomic mass of Ta + (7 * atomic mass of F)
= (2 * 39.1 g/mol) + 180.95 g/mol + (7 * 19.0 g/mol)
= 391.1 g/mol + 180.95 g/mol + 133.0 g/mol
= 705.05 g/mol
Moles of K2TaF7 = mass of K2TaF7 / molar mass of K2TaF7
= 208.0 kg * (1000 g/kg) / 705.05 g/mol
= 295.19 mol
Molar mass of Na: 22.99 g/mol
Moles of Na = mass of Na / molar mass of Na
= 28.0 kg * (1000 g/kg) / 22.99 g/mol
= 1217.48 mol
Now, let's determine how many moles of each substance we have based on the stoichiometry:
Moles of Ta = 7 * (moles of K2TaF7 / 2)
= 7 * (295.19 mol / 2)
= 1029.17 mol
Moles of NaF = 14 * (moles of Na / 2)
= 14 * (1217.48 mol / 2)
= 8512.76 mol
Moles of KF = 4 * (moles of K2TaF7 / 2)
= 4 * (295.19 mol / 2)
= 590.38 mol
Now, let's convert the moles of each substance back to kilograms:
Mass of K2TaF7 remaining = (moles of K2TaF7 - moles of Ta) * molar mass of K2TaF7
= (295.19 mol - 1029.17 mol) * 705.05 g/mol / (1000 g/kg)
= -547.97 kg (Note: The negative sign indicates that all the K2TaF7 is used up)
Mass of Na remaining = (moles of Na - moles of NaF) * molar mass of Na
= (1217.48 mol - 8512.76 mol) * 22.99 g/mol / (1000 g/kg)
= -172.67 kg (Note: The negative sign indicates that all the Na is used up)
Mass of Ta = moles of Ta * molar mass of Ta
= 1029.17 mol * 180.95 g/mol / (1000 g/kg)
= 186.51 kg
Mass of KF = moles of KF * molar mass of KF
= 590.38 mol * 58.1 g/mol / (1000 g/kg)
= 34.26 kg
Mass of NaF = moles of NaF * molar mass of NaF
= 8512.76 mol * 41.99 g/mol / (1000 g/kg)
= 357.17 kg
Therefore, the reactor contents at the completion of this reaction will be:
a) The number of kg of K2TaF7 remaining: -547.97 kg
b) The number of kg of Na remaining: -172.67 kg
c) The number of kg of Ta present: 186.51 kg
d) The number of kg of KF present: 34.26 kg
e) The number of kg of NaF present: 357.17 kg