Final answer:
From 6000 kg of ulexite ore with an 89% yield and 0.032% ulexite by mass, approximately 279.91 grams of boron can be produced.
Step-by-step explanation:
To calculate how many grams of boron can be produced from 6000 kg of ulexite-bearing ore with a composition of 0.032% ulexite by mass and an 89% yield, we first need to determine the mass of ulexite in the ore. Then, we can calculate the mass of boron that can be extracted.
Firstly, we find the mass of ulexite:
- Mass of ulexite = 6000 kg × 0.032% = 6000 × 0.00032 = 1.92 kg
Next, we determine how many moles of ulexite this mass represents. The molar mass of ulexite, NaCaB5O9·•2H2O, can be calculated from the atomic masses of its constituent elements (sodium, calcium, boron, oxygen, and hydrogen).
Assuming we have calculated the molar mass to be approximately 330 g/mol, we then convert the mass of ulexite to moles:
- Moles of ulexite = 1920 g / 330 g/mol ≈ 5.818 moles
Since each mole of ulexite contains 5 moles of boron, the moles of boron are:
- Moles of boron = 5.818 moles × 5 = 29.09 moles
Now, we find the mass of boron:
- Mass of boron = 29.09 moles × 10.81 g/mol (approx average atomic mass of boron) = 314.50 g
Finally, we apply the process yield to find the actual mass of boron produced:
- Actual mass of boron = 314.50 g × 89% = 279.91 g
Therefore, from 6000 kg of ulexite ore with an 89% yield and 0.032% ulexite by mass, we can produce approximately 279.91 grams of boron.