Hướng dẫn:

Gọi a, b lần lượt là nồng độ mol của $Cu{{\left( N{{O}_{3}} \right)}_{2}},AgN{{O}_{3}}$

${{n}_{Cu{{\left( N{{O}_{3}} \right)}_{2}}}}=V.{{C}_{{{M}_{Cu{{\left( N{{O}_{3}} \right)}_{2}}}}}}=0,2.a\left( mol \right)$ 

${{n}_{AgN{{O}_{3}}}}=V.{{C}_{{{M}_{_{AgN{{O}_{3}}}}}}}=0,2.b\left( mol \right)$

  $Fe+Cu{{\left( N{{O}_{3}} \right)}_{2}}\to Fe{{\left( N{{O}_{3}} \right)}_{2}}+Cu$

0,2.a $\leftarrow$ 0,2.a      $\to$                   0,2.a           (mol)

  $Fe+2AgN{{O}_{3}}\to Fe{{\left( N{{O}_{3}} \right)}_{2}}+2Ag$

.0,1.b$\leftarrow $0,2.b   $\to $                      0,2.b       (mol)

Theo phương trình ta có:

${{n}_{Fe}}={{n}_{Cu{{\left( N{{O}_{3}} \right)}_{2}}}}+\frac{{{n}_{AgN{{O}_{3}}}}}{2}=0,2.a+\frac{0,2.b}{2}=0,2.a+0,1.b$ 

$\Rightarrow 0,2.a+0,1.b=0,125\left( 1 \right)$ 

${{n}_{Cu}}={{n}_{Cu{{\left( N{{O}_{3}} \right)}_{2}}}}=0,2.a\left( mol \right)$ 

${{n}_{Ag}}={{n}_{AgN{{O}_{3}}}}=0,2.b\left( mol \right)$ 

$\Rightarrow {{m}_{KL}}={{m}_{Cu}}+{{m}_{Ag}}={{n}_{Cu}}.{{M}_{Cu}}+{{n}_{Ag}}.{{M}_{Ag}}$ 

$\Rightarrow 64.0,2.a+108.0,2.b=15,6$ 

$\Rightarrow 12,8.a+21,6.b=15,6\left( 2 \right)$

Từ (1) và (2) có: $\begin{align}& a=0,375 \\ & b=0,5 \\ \end{align}$ (M)

 Chọn đáp án B.