Answer:
34.44 kg
Explanation:
First we convert 1.00 kg of phosphorus (P) into moles, using its molar mass:
- 1.00 kg ÷ 32 kg/kmol = 0.03125 kmol P
Then we convert 0.03125 kmoles of P into kmoles of Ca₃(PO₄)₂:
- 0.03125 kmol P * [tex]\frac{2kmolCa_3(PO_4)_2}{2kmolP}[/tex] = 0.0625 kmol Ca₃(PO₄)₂
Now we calculate the mass of 0.0625 kmoles of Ca₃(PO₄)₂:
- 0.0625 kmol Ca₃(PO₄)₂ * 310.18 kg/kmol = 19.39 kg
Finally we calculate the required mass of the ore, using the definition of content percentage:
- % content = Mass of calcium phosphate / mass of ore * 100 %
- 56.3 % = 19.39 kg / mass of ore * 100%
