Answer:
[tex]\boxed{\textbf{Mass of ore = 26.7 kg}}[/tex]
Explanation:
1. Calculate the mass of Ca₃(PO₄)₂
1 mol (310.18 g) of Ca₃(PO₄)₂ contains 61.95 g of P
[tex]\text{Mass of Ca$_{3}$(PO$_{4}$)$_{2}$} = \text{3.53 kg P} \times \dfrac{\text{310.18 kg Ca$_{3}$(PO$_{4}$)$_{2}$}}{\text{61.95 kg P}}\\\\= \text{17.67 kg Ca$_{3}$(PO$_{4}$)$_{2}$}[/tex]
2. Calculate the mass of ore
[tex]\text{Mass of ore} = \text{17.67 kg Ca$_{3}$(PO4$_{4}$)$_{2}$} \times \dfrac{\text{100 kg ore}}{\text{66.1kg Ca$_{3}$(PO4$_{4}$)$_{2}$} } = \textbf{26.7 kg ore}\\\\\boxed{\textbf{Mass of ore = 26.7 kg}}[/tex]
