Помогите решить задачу до завтра пожалуйста!!!

дано

m(P) = 1.55 g

V ppa(Ca(OH)2) = 3.7 L

p(Ca(OH)2) = 1 g/mL

W(Ca(OH)2) = 0.1%

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m(Ca3(PO4)2)-?

4P+5O2--gt;2P2O5

M(P) = 31 g/mol

n(P) = m/M = 1.55 / 31 = 0.05 mol

4n(P) = 2n(P2O5)

n(P2O5) = 2*0.05/4 = 0.025 mol

M(P2O5) = 142 g/mol

m(P2O5) = n*M = 0.025*142 = 3.55 g

P2O5+3Ca(OH)2--gt;Ca3(PO4)2+3H2O

m(Ca(OH)2)= 3.7*1*0.1 %/ 100% = 0.0037g

n(P2O5)= m(P2O5) / M(P2O5) = 3.55 / 142 = 0.0025 mol

M(Ca(OH)2)= 74 g/mol

n(Ca(OH)2) = m/M= 0.0037 / 74  = 0.00005 mol

n(P2O5) gt; n(Ca(OH)2)

3n(Ca(OH)2) = n(Ca3(PO4)2)

n(Ca3(PO4)2)= 0.00005 / 3 = 0.000017 mol

M(Ca3(PO4)2) = 310 g/mol

m(Ca3(PO4)2) = n*M = 0.00017 * 310 = 0.005 g

ответ 0.005 г