# Answer to Question #33008 in Other Chemistry for Rushdiya

Question #33008

Ca3(PO4)2 + 4H3PO4 -- 3Ca (H2OP4)4
calculate the required mass of Ca3(PO4)2 to obtain
100g of Ca (H2OP4)4

Expert's answer

Ca

m of Ca(H

First of all we can find the amount of substance of Ca(H

n = m/M

where

n - the amount of substance

m - the mass

M - the molar mass, which is calculated: M(Ca(H

n(Ca(H

According the equation Ca

n(Ca

M(Ca

m(Ca

Answer: 12.4g

_{3}(PO_{4})_{2}+ 4H_{3}PO_{4}-- 3Ca(H_{2}PO_{4})_{2}m of Ca(H

_{2}PO_{4})_{2}is 100gFirst of all we can find the amount of substance of Ca(H

_{2}PO_{4})_{2}. We can use the formulan = m/M

where

n - the amount of substance

m - the mass

M - the molar mass, which is calculated: M(Ca(H

_{2}PO_{4})_{2}) = 40 +4*2*1+4*16+31*4*4 = 31+ 3 *1 = 608 g/moln(Ca(H

_{2}PO_{4})_{2}) = 100 g / 608 g/mol = 0.164 molesAccording the equation Ca

_{3}(PO4)_{2}+ 4H_{3}PO_{4}--> 3Ca(H_{2}PO_{4})_{2}we can find the amount of substance of Ca_{3}(PO_{4})_{2}:n(Ca

_{3}(PO_{4})_{2}) = n(Ca(H_{2}PO_{4})_{2})/3 = 0.164/4 = 0.04 molesM(Ca

_{3}(PO_{4})_{2}= 310 g/molm(Ca

_{3}(PO_{4})_{2}) = 0.04moles * 310 g/mol = 12.4gAnswer: 12.4g

## Comments

## Leave a comment