Answer to Question #33008 in Chemistry for Rushdiya

Question

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

Answer 1

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