Question

Phosphorus is obtained primarily from ores containing calcium phosphate. part a if a particular ore contains 55.9 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosphorus? express your answer with the appropriate units.

Answer 1

The percentage of calcium phosphate in ore = 55. 9% so it means each 100g of ore contains 55.9 g of calcium phosphate

The formula of calcium phosphate = Ca3(PO4)3

As per molecular formula each mole of calcium phosphate contains three moles of phosphorous

mass of each mole of calcium phosphate = 310 g

for 31 g of P we need = 310 /3 g of calcium phosphate  

                                     = 103.33 g of calcium phosphate  

for 1 g of P we need = 103.33 / 31 g of calcium phosphate = 3.33 g

So for 1000g of P we need = 3.33 X 1000g  of calcium phosphate

                                                 = 3333.3 g of calcium phosphate

now for 55.9 g of Calcium phosphate we need = 100 g of ore

so for 3333.3  g of calcium phosphate we need = 100 X 3333.3 / 55.9 g

                                                                                    = 5963.03

Answer 2

Answer:

$m_{ore}=8.94kgOre$

Explanation:

Hello,

To know the minimum mass of the ore, one must apply the following mole-mass relationship in which we consider there are two phosphorous atoms into the calcium phosphate and the 55.9% purity as a division since the pure calcium phosphate is just a fraction of the whole ore, taking into account that the calcium phosphate has the following formula:

$Ca_3(PO_4)_2$

$m_{ore}=1.00kg*\frac{1000gP}{1kgP}*\frac{1molP}{31gP}*\frac{1molCa_3(PO_4)_2}{2molP}*\frac{310gCa_3(PO_4)_2}{1molCa_3(PO_4)_2}*\frac{100gCa_3(PO_4)_2}{55.9gCa_3(PO_4)_2}\\m_{ore}=8944.54gOre\\m_{ore}=8.94kgOre$

Best regards.