Question

Phosphorus is obtained primarily from ores containing calcium phosphate. If a particular ore contains 66.1% calcium phosphate, what minimum mass of the ore must be processed to obtain 3.53 kg of phosphorus?

Answer 1

`\boxed{\textbf{Mass of ore = 26.7 kg}}`

Step-by-step explanation:

1. Calculate the mass of Ca₃(PO₄)₂

1 mol (310.18 g) of Ca₃(PO₄)₂ contains 61.95 g of P

`\text{Mass of Ca$_(3)$(PO$_(4)$)$_(2)$} = \text{3.53 kg P} * \frac{\text{310.18 kg Ca$_(3)$(PO$_(4)$)$_(2)$}}{\text{61.95 kg P}}\\\\= \text{17.67 kg Ca$_(3)$(PO$_(4)$)$_(2)$}`

2. Calculate the mass of ore

`\text{Mass of ore} = \text{17.67 kg Ca$_(3)$(PO4$_(4)$)$_(2)$} * \frac{\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}}`