Question

In a sample of calcium phosphate 0.432
mole of phosphorous is present, what is amount of
calcium phosphate is present in the sample if the
sample is 100% pure? [Ca = 40, P = 31, 0 = 16)​

Answer 1

Answer : The amount of calcium phosphate is present in the sample is 66.96 grams.

Explanation :

The formula of calcium phosphate is, $Ca_3(PO_4)_2$

In $Ca_3(PO_4)_2$ compound, there are 3 moles of calcium atom, 2 moles of phosphorous atom and 8 moles of oxygen atoms.

First we have to calculate the moles of $Ca_3(PO_4)_2$.

As, 2 moles of phosphorous present in 1 mole of $Ca_3(PO_4)_2$

So, 0.432 moles of phosphorous present in $\frac{0.432}{2}=0.216$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $Ca_3(PO_4)_2$.

Molar mass of $Ca_3(PO_4)_2$ = $3(40g/mol)+2(31g/mol)+8(16g/mol)=310g/mol$

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.216moles)\times (310g/mole)=66.96g$

Therefore, the amount of calcium phosphate is present in the sample is 66.96 grams.

Answer 2

Answer:

the answer is 67(approx)gm.

Explanation:

see the above attachment.