Đáp án:

\({m_{dd{\text{ }}B}} = 261{\text{ }}gam\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(F{\text{e}} + {H_2}S{O_4}\xrightarrow{{}}F{\text{eS}}{{\text{O}}_4} + {H_2}\)

\(2{\text{A}}l + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)

Gọi số mol \(Al;Fe\) trong hỗn hợp lần lượt là \(x;y\)

Ta có:

\({m_{Al}} = 27{\text{x}};{m_{F{\text{e}}}} = 56y\)

\( \to {m_{Al}}:{m_{F{\text{e}}}} = 27{\text{x}}:56y = 9:28 \to \frac{{27{\text{x}}}}{{56y}} = \frac{9}{{28}} \to 3{\text{x}} = 2y\)

 Ta có:

\({n_{{H_2}}} = \frac{{13,44}}{{22,4}} = 0,6{\text{ mol = }}{{\text{n}}_{{H_2}S{O_4}}} = \frac{3}{2}{n_{Al}} + {n_{F{\text{e}}}} = 1,5{\text{x}} + y\)

 Giải được: \(x=0,2;y=0,3\)

\({m_{{H_2}S{O_4}}} = 0,6.98 = 58,8{\text{ gam}} \to {{\text{m}}_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{58,8}}{{24,5\% }} = 240{\text{ gam}}\)

BTKL:

\({m_{Al}} + {m_{F{\text{e}}}} + {m_{dd\;{{\text{H}}_2}S{O_4}}} = {m_{dd\;{\text{B}}}} + {m_{{H_2}}}\)

\( \to 0,2.27 + 0,3.56 + 240 = {m_{dd\;{\text{B}}}} + 0,6.2\)                                                      

\( \to {m_{dd{\text{ }}B}} = 261{\text{ }}gam\)

Đặt $n_{Al}=x(mol);n_{Fe}=y(mol)$

$\Rightarrow m_{Al}=27x(g);m_{Fe}=56y(g)\\ \Rightarrow m_{Al}:m_{Al}=27x:56y=9:28\\ \Rightarrow x=\dfrac{504y}{756}=\dfrac{2}{3}y(1)\\ PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow 1,5x+y=n_{H_2}=\dfrac{13,44}{22,4}=0,6(2)\\ (1)(2)\Rightarrow x=0,2(mol);y=0,3(mol)\\ \Rightarrow n_{FeSO_4}=y=0,3(mol);n_{Al_2(SO_4)_3}=0,5.x=0,1(mol)\\ n_{H_2SO_4}=n_{H_2}=0,6(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,6.98}{24,5\%}=240(g)\\m_{FeSO_4}=0,3.152=45,6(g);m_{Al_2(SO_4)_3}=0,1.342=34,2(g)\\ m_{hh}=m_{Fe}+m_{Al}=0,2.27+0,3.56=22,2(g);m_{H_2}=0,6.2=1,2(g)\\ \Rightarrow m_{dd B}=22,2+240-1,2=261(g)$