дано
m(ppa CaCL2) = 44.4 g
W(CaCL2) = 15%
+H3PO4
m(Ca3(PO4)2-?
m(CaCL2) = 44.4*15% / 100% = 6.66 g
3CaCL2+2H3PO4-->Ca3(PO4)2↓+6HCL
M(CaCL2) = 111 g/mol
n(CaCL2) = m/M = 6.66 / 111=0.06 mol
3n(CaCL2) = n(Ca3(PO4)2
n(Ca3(PO4)2) = 0.06 / 3 = 0.02 mol
M(Ca3(PO4)2) = 310 g/mol
m(Ca3(PO4)2 * n*M = 0.02 * 310 = 6.2 g
ответ 6.2 г
дано
m(ppa CaCL2) = 44.4 g
W(CaCL2) = 15%
+H3PO4
m(Ca3(PO4)2-?
m(CaCL2) = 44.4*15% / 100% = 6.66 g
3CaCL2+2H3PO4-->Ca3(PO4)2↓+6HCL
M(CaCL2) = 111 g/mol
n(CaCL2) = m/M = 6.66 / 111=0.06 mol
3n(CaCL2) = n(Ca3(PO4)2
n(Ca3(PO4)2) = 0.06 / 3 = 0.02 mol
M(Ca3(PO4)2) = 310 g/mol
m(Ca3(PO4)2 * n*M = 0.02 * 310 = 6.2 g
ответ 6.2 г