Đáp án:

$Fe+6HNO_3→Fe(NO_3)_3+3NO_2+H_2O$

   1/3a       2a            a              3a 

$Cu+4HNO_3→Cu(NO_3)_2+2NO_2+2H_2O$

    1/2b      2b             b             2b

$a)n_{NO_2}$=$\dfrac{17,92}{22,4}=0,8(mol)$

Ta có hpt $\left \{ {{a+b=0,8} \atop {{\dfrac{56a}3+32b=17,6}}} \right.$ $⇒a=0,6(mol); b=0,2(mol)$

$⇒n_{Fe}=1/3.0,6=0,2(mol); n_{Cu}=1/2.0,2=0,1(mol)$

$⇒$%$_{F_e}=$$\dfrac{m_{Fe}.100}{m_{h^2}}=$ $\dfrac{56.0,2}{17,6}.100≈64$ %

$⇒$ %$Cu=36$%

$b)n_{HNO_3}=2.0,6+2.0,2=1,6(mol)$

$C_M=$$\dfrac{1,6}{0,5}=3,2(M)$

BẠN THAM KHẢO NHA!!!

Đáp án:

\(\begin{array}{l}
a)\\
\% Fe = 63,64\% \\
\% Cu = 36,36\% \\
b)\\
{C_{{M_{HN{O_3}}}}} = 3,2M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Fe + 6HN{O_3} \to Fe{(NO)_3} + 3N{O_2} + 3{H_2}O\\
Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
{n_{N{O_2}}} = \dfrac{{17,92}}{{22,4}} = 0,8mol\\
hh:Fe(a\,mol),Cu(b\,mol)\\
3a + 2b = 0,8(1)\\
56a + 64b = 17,6(2)\\
\text{Từ (1) và (2)} \Rightarrow a = 0,2;b = 0,1\\
{m_{Fe}} = 0,2 \times 56 = 11,2g\\
{m_{Cu}} = 17,6 - 11,2 = 6,4g\\
\% Fe = \dfrac{{11,2}}{{17,6}} \times 100\%  = 63,64\% \\
\% Cu = 100 - 63,64 = 36,36\% \\
b)\\
{n_{HN{O_3}}} = 6{n_{Fe}} + 4{n_{Cu}} = 6 \times 0,2 + 4 \times 0,1 = 1,6mol\\
{C_{{M_{HN{O_3}}}}} = \dfrac{n}{V} = \dfrac{{1,6}}{{0,5}} = 3,2M
\end{array}\)