Дано
m(ppa CaCL2) = 166.5 g
W(CaCL2) = 20%
+ Na3PO4
--------------------
m(Ca3(PO4)2)-?
m(CaCL2) = m(ppa CaCL2) * W(CaCL2) / 100%
m(CaCL2)= 166.5 * 20% / 100% = 33.3 g
33.3g X g
3CaCL2+2Na3PO4-->Ca3(PO4)2+6NaCL
3*111 310
M(CaCL2) = 111 g/mol , M(Ca3(PO4)2) = 310 g/mol
X= 33.3 * 310 / 333 = 31 g
ответ 31 г