Hướng dẫn:

$\begin{align}  & {{n}_{CaO}}=\frac{{{m}_{CaO}}}{{{M}_{CaO}}}=\frac{1,68}{56}=0,03mol \\ & CaO+{{H}_{2}}O\to Ca{{(OH)}_{2}} \\ & 0,03\to 0,03\to 0,03mol \\ & T=\frac{{{n}_{O{{H}^{-}}}}}{{{n}_{{{H}_{3}}P{{O}_{4}}}}}=\frac{2{{n}_{Ca{{(OH)}_{2}}}}}{{{n}_{{{H}_{3}}P{{O}_{4}}}}}=\frac{2.0,03}{0,02}=3 \\\end{align}$

$\Rightarrow $ $2{{H}_{3}}P{{O}_{4}}+3Ca{{(OH)}_{2}}\to C{{a}_{3}}{{(P{{O}_{4}})}_{2}}+6{{H}_{2}}O$

      0,02$\to $                     0,01 mol

${{m}_{C{{a}_{3}}{{(P{{O}_{4}})}_{2}}}}={{n}_{C{{a}_{3}}{{(P{{O}_{4}})}_{2}}}}.{{M}_{C{{a}_{3}}{{(P{{O}_{4}})}_{2}}}}=0,01.310=3,1g$

                   $\Rightarrow $ Chọn đáp án A.