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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
Ca3(PO4)2 + 4H3PO4 -- 3Ca(H2PO4)2

m of Ca(H2PO4)2 is 100g

First of all we can find the amount of substance of Ca(H2PO4)2. We can use the formula
n = m/M
where
n - the amount of substance
m - the mass
M - the molar mass, which is calculated: M(Ca(H2PO4)2) = 40 +4*2*1+4*16+31*4*4 = 31+ 3 *1 = 608 g/mol
n(Ca(H2PO4)2) = 100 g / 608 g/mol = 0.164 moles
According the equation Ca3(PO4)2 + 4H3PO4 --> 3Ca(H2PO4)2 we can find the amount of substance of Ca3(PO4)2:
n(Ca3(PO4)2) = n(Ca(H2PO4)2)/3 = 0.164/4 = 0.04 moles
M(Ca3(PO4)2 = 310 g/mol
m(Ca3(PO4)2) = 0.04moles * 310 g/mol = 12.4g
Answer: 12.4g

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