Ответ:
дано
m(ppa CaCL2) = 130 g
W(CaCL2) = 5%
--------------------
m(Ca3(PO4)2)-?
m(CaCL2) = 130 * 5% / 100% = 6.5 g
3CaCL2+3H3PO4-->Ca3(PO4)2+6HCL
M(CaCL2) =111 g/mol
n(CaCL2) = m/M = 6.5 / 111 = 0.06 mol
n(CaCL2) = n(Ca3(PO4)2)
M(Ca3(PO4)2)= 310 g/mol
m(Ca3(PO4)2) = n*M = 0.06 * 310 = 18.6 g
ответ 18.6 г
Объяснение:
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Ответ:
дано
m(ppa CaCL2) = 130 g
W(CaCL2) = 5%
--------------------
m(Ca3(PO4)2)-?
m(CaCL2) = 130 * 5% / 100% = 6.5 g
3CaCL2+3H3PO4-->Ca3(PO4)2+6HCL
M(CaCL2) =111 g/mol
n(CaCL2) = m/M = 6.5 / 111 = 0.06 mol
n(CaCL2) = n(Ca3(PO4)2)
M(Ca3(PO4)2)= 310 g/mol
m(Ca3(PO4)2) = n*M = 0.06 * 310 = 18.6 g
ответ 18.6 г
Объяснение: