Đáp án:
$ \left\{\begin{matrix}\%m_{Al}=36\%\\\%m_{Cu}=64\%\end{matrix}\right.$
$ m_{dd\ H_2SO_4}=165\ (g)$
$ m_{\text{muối}}=76,56\ (g)$
Giải thích các bước giải:
a,
Đặt: $\left\{\begin{matrix}n_{Al}=x\ (mol)\\n_{Cu}=y\ (mol)\end{matrix}\right.$
$\to 27x+64y=15\ (1)$
$n_{SO_2}=\dfrac{10,08}{22,4}=0,45\ (mol)$
BTe:
$\mathop{Al}\limits^0-3e\to \mathop{Al}\limits^{+3}\qquad ||\qquad \mathop{S}\limits^{+6}+2e\to \mathop{S}\limits^{+4}\\\ x\ \quad3x \qquad \qquad \qquad \qquad \quad 0,9\quad 0,45\\\mathop{Cu}\limits^0-2e\to \mathop{Cu}\limits^{+2}\\\ y\quad \ \ \ 2y$
$\to 3x+2y=0,9\ (2)$
Từ $(1)$ và $(2)$ giải hệ ta được: $\left\{\begin{matrix}x=0,2\ (mol)\\y=0,15\ (mol)\end{matrix}\right.$
$\to \left\{\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{15}.100\%=36\%\\\%m_{Cu}=100\%-36\%=64\%\end{matrix}\right.$
b,
$H_2SO_4+2NaOH\to Na_2SO_4+2H_2O$
$n_{NaOH}=0,5.3=1,5\ (mol)$
$\to n_{H_2SO_4\ \text{dư}}=\dfrac12 .1,5=0,75\ (mol)$
BTNT $S$, ta có:
$n_{H_2SO_4\ \text{pứ}}=3n_{Al_2(SO_4)_3}+n_{CuSO_4}+n_{SO_2}$
$=3.\dfrac12 .0,2+0,15+0,45=0,9\ (mol)$
$\to \sum n_{H_2SO_4}=0,75+0,9=1,65\ (mol)$
$\to m_{dd\ H_2SO_4}=\dfrac{1,65.98.100}{98}=165\ (g)$
c,
Đặt $n_{Fe_2S}=x\ (mol)\to n_{Cu_2S}=3x\ (mol)$
$2FeS_2+14H_2SO_4\to Fe_2(SO_4)_3+15SO_2+14H_2O\ (1)$
$Cu_2S+6H_2SO_4\to 2CuSO_4+5SO_2+6H_2O\ (2)$
Theo PT $(1):\ n_{H_2SO_4}=7n_{FeS_2}=7x$
Theo PT $(2):\ n_{H_2SO_4}=6n_{Cu_2S}=18x$
$\to \sum n_{H_2SO_4}=25x=1,65\ (mol)$
$\to x=0,066\ (mol)$
Theo PT $(1):\ n_{Fe_2(SO_4)_3}=\dfrac12.0,066=0,033\ (mol)$
Theo PT $(2):\ n_{CuSO_4}=2n_{Cu_2S}=2.3.0,066=0,396\ (mol)$
$\to m_{\text{muối}}=0,033.400+0,396.160=76,56\ (g)$
