Question

Đã trả lời bởi chuyên gia

Answer 1

\({n_{A{s_4}{O_6}}} = \frac{{0,1188}}{{75 \cdot 4 + 16 \cdot 6}} = {3.10^{ - 4}}\,{\rm{mol}}{\rm{. }}\)

Phương trình hóa học:

\(\begin{array}{l}{\rm{A}}{{\rm{s}}_4}{{\rm{O}}_6} + 6{{\rm{H}}_2}{\rm{O}} \to 4{{\rm{H}}_3}{\rm{As}}{{\rm{O}}_3}\\{3.10^{ - 4}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,0012\,\,\,\,\,\,\,\,\,\,mol\end{array}\)

\(\begin{array}{l}{{\rm{H}}_3}{\rm{As}}{{\rm{O}}_3}(aq) + I_3^ - \,(aq) + {{\rm{H}}_2}{\rm{O}}\,(l) \to {{\rm{H}}_3}{\rm{As}}{{\rm{O}}_4}(aq) + 3{I^ - }(aq) + 2{{\rm{H}}^ + }(aq)\\0,0012\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,0012\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol\end{array}\)

\({C_{I_3^ - }} = \frac{{0,0012}}{{0,01}} = 0,12M\)

Lượng \(I_3^ - \)tham gia phản ứng với \({\rm{HCN}} = 0,005 \cdot 0,12 = {6.10^{ - 4}}\) (mol)

\(\begin{array}{l}HCN\,(aq) + I_3^ - \,(aq) \to ICN\,(aq) + 2{I^ - }(aq) + {H^ + }(aq)\\{6.10^{ - 4}}\,\,\,\,\,\,\, \leftarrow \,\,{6.10^{ - 4}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol\end{array}\)

Nồng độ mol của \({\rm{HCN}}\) trong mẫu máu là \({C_{HCN}} = \frac{{6 \cdot {{10}^{ - 4}}}}{{0,015}} = 0,04M.\)

Chọn A.

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Answer 2