:v ghê như thật

\(n_{Al}=\dfrac{10,8}{27}=0,4(mol)\\ a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,6(mol);n_{Al_2(SO_4)_3}=0,2(mol)\\ b,V_{H_2}=0,6.22,4=13,44(l)\\ c,m_{dd_{H_2SO_4}}=\dfrac{0,6.98}{9,8\%}=600(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,2.342}{10,8+600-0,6.2}.100\%=11,22\%\)

\(Cu+2AgNO_3\to Cu(NO_3)_2+2Ag\\ \Rightarrow n_{Cu}=\dfrac{1}{2}n_{AgNO_3}=\dfrac{1}{2}.0,2.0,1=0,01(mol)\\ \Rightarrow m_{Cu}=0,01.64=0,64(g)\)

\(Fe+CuSO_4\to FeSO_4+Cu\)

Đặt \(n_{CuSO_4}=x(mol)\)

\(\Rightarrow 64x-56x=2\\ \Rightarrow x=0,25(mol)\\ \Rightarrow n_{Fe}=n_{Cu}=0,25(mol)\\ \Rightarrow m_{Fe}=0,25.56=14(g);m_{Cu}=0,25.64=16(g)\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=0,2(mol)\\ a,V_{H_2}=0,2.22,4=4,48(l)\\ b,m_{dd_{H_2SO_4}}=\dfrac{0,2.98}{9,8\%}=200(g)\\ c,C\%_{FeSO_4}=\dfrac{0,2.152}{11,2+200-0,2.2}.100\%=14,42\%\)

\(\Leftrightarrow3x+10=25\\ \Leftrightarrow3x=25-10=15\\ \Leftrightarrow x=\dfrac{15}{3}=5\)

\(b,\Leftrightarrow\left[{}\begin{matrix}9-7x=5x-3\left(x\le\dfrac{9}{7}\right)\\7x-9=5x-3\left(x>\dfrac{9}{7}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\\ c,\Leftrightarrow\dfrac{12x}{60}=\dfrac{15y}{60}=\dfrac{20z}{60}\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y+z}{5+4+3}=\dfrac{48}{12}=4\\ \Leftrightarrow\left\{{}\begin{matrix}x=20\\y=16\\z=12\end{matrix}\right.\)

Bài 2:

\(a,\Leftrightarrow x-1=-2\Leftrightarrow x=-1\\ c,\Leftrightarrow\sqrt{x}\left(\sqrt{x}-3\right)=0\left(x\ge0\right)\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\left(tm\right)\)

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