Final answer:
To find the minimum mass of the ore required to obtain 4.13 kg of phosphorus, calculate the molar mass of calcium phosphate, determine the moles of phosphorus in 4.13 kg, and consider that the ore is 35.1% calcium phosphate. Divide the mass of calcium phosphate required by this percentage to determine the total mass of the ore needed.
Step-by-step explanation:
To calculate the minimum mass of ore needed to obtain 4.13 kg of phosphorus, we must first consider the chemical composition of the ore. The ore contains calcium phosphate, which has a formula of Ca3(PO4)2. Since we are given that the ore is 35.1% calcium phosphate, we can calculate the amount of calcium phosphate that will yield the desired amount of phosphorus.
First, we need to determine the molar mass of calcium phosphate: Ca3(PO4)2, which consists of 3 calcium atoms, 2 phosphorus atoms, and 8 oxygen atoms. The molar masses are approximately 40 g/mol for calcium, 31 g/mol for phosphorus, and 16 g/mol for oxygen, leading to a total molar mass of 310 g/mol for calcium phosphate (3*40 + 2*31 + 8*16). Next, we calculate how many moles of phosphorus there are in 4.13 kg, considering the molecular ratio in calcium phosphate (1 mole of calcium phosphate contains 2 moles of phosphorus).
Since the ore is only 35.1% calcium phosphate, we then divide the mass of calcium phosphate needed by this percentage to find the total mass of the ore required. This will give us the minimum mass of the ore that must be processed to obtain 4.13 kg of phosphorus.