Question

Answer 1

Câu 10: 

a, Gọi: \(\left\{{}\begin{matrix}n_{Cu}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

⇒ 64x + 56y = 2,4 (1)

Ta có: n_NO2 = 0,1 (mol)

Theo ĐLBT e, có: 2x + 3y = 0,1 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,02\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,02.64}{2,4}.100\%\approx53,33\%\\\%m_{Fe}\approx46,67\%\end{matrix}\right.\)

b, Ta có: n_HNO3 = 2n_NO2 = 0,2 (mol)

\(\Rightarrow m_{ddHNO_3}=\dfrac{0,2.63}{60\%}=21\left(g\right)\)

c, BTNT Cu: n_Cu(NO3)2 = n_Cu = 0,02 (mol)

BTNT Fe, có: n_Fe(NO3)3 = n_Fe = 0,02 (mol)

Ta có: m _{dd sau pư} = 2,4 + 21 - 0,1.46 = 18,8 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Cu\left(NO_3\right)_2}=\dfrac{0,02.188}{18,8}.100\%=20\%\\C\%_{Fe\left(NO_3\right)_3}=\dfrac{0,02.242}{18,8}.100\%\approx25,74\%\end{matrix}\right.\)