\(n_{Al}=\dfrac{3.5}{27}=\dfrac{7}{54}\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{180\cdot12.25\%}{98}=0.225\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(2.........3\)

\(\dfrac{7}{54}.....0.225\)

\(LTL:\dfrac{\dfrac{7}{54}}{2}< \dfrac{0.225}{3}\Rightarrow H_2SO_4dư\)

\(n_{H_2}=\dfrac{7}{54}\cdot\dfrac{3}{2}=\dfrac{7}{36}\left(mol\right)\)

\(V_{H_2}=\dfrac{7}{36}\cdot22.4=4336\left(l\right)\)

\(m_{dd}=3.5+180-\dfrac{7}{36}\cdot2=183.11\left(g\right)\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{\dfrac{7}{108}\cdot342}{183.11}\cdot100\%=12.1\%\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0.225-\dfrac{7}{36}\right)\cdot98}{183.11}\cdot100\%=1.63\%\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot29,4\%}{98}=0,6\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,6}{3}\) \(\Rightarrow\) Axit còn dư, Nhôm p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,6-0,3=0,3\left(mol\right)\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,3\cdot98=29,4\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72 \left(l\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Al}+m_{ddH_2SO_4}-m_{H_2}=204,8\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{204,8}\cdot100\%\approx16,7\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{29,4}{204,8}\cdot100\%\approx14,36\%\end{matrix}\right.\)

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

0,4           0,3                 0,2        ( mol )

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\) ( không giải phóng H2)

=> Chất rắn tạo ra H2 là Al

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

 0,2                                                 0,3     ( mol )

\(\Rightarrow m_{Al}=\left(0,2.27\right)+\left(0,4.27\right)=16,2g\)

a)

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

Theo PTHH : 

\(n_{CuSO_4} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)\\ \Rightarrow V_{dd\ CuSO_4} = \dfrac{0,2}{0,1} = 2(lít)\)

b)

\(n_{CuSO_4} = 0,1.2,5 = 0,25(mol) > n_{Fe}= 0,2\) nên \(CuSO_4\) dư.

Ta có : 

\(n_{CuSO_4\ pư} = n_{FeSO_4} = n_{Fe} = 0,2(mol)\\ \Rightarrow n_{CuSO_4\ dư} = 0,25 - 0,2 = 0,05(mol)\)

Vậy : 

\(C_{M_{CuSO_4}} = \dfrac{0,05}{2,5} = 0,02M\\ C_{M_{FeSO_4}} = \dfrac{0,2}{2,5} = 0,08M\)

3,5 gam gì vậy

Đáp án A

PTHH: 2Al  +  3H₂SO₄   --->   Al₂(SO₄)₃  +  3H₂

Ag không phản ứng với HCl vì trong dãy hoạt động hóa học thì Ag đứng sau H.

=> \(n_{H_2}=\frac{13,44}{22,4}=0,6\) (mol)

Theo PTHH: \(n_{Al}=\frac{2}{3}n_{H_2}=\frac{2}{3}.0,6=0,4\) (mol)

=> m_Al = 0,4.27 = 10,8 (g)

=> m_Ag = 12 - 10,8 = 1,2(g)

a)PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

(mol) 2 3 1 3

(mol) 0,4 <-- 0,6 <-- 0,2 <-- 0,6

n_H2= V/22,4 = 13,44/22,4 = 0,6(mol)

m_Al=n.M= 0,4.27 = 10,8(g)

%m_{Al =} m_Al/m_hh.100% = 10,8/12.100% = 90%

b) m_{H2SO4 =} n.M = 0,6.98=58,8(g)

m_{ddH2SO4 =} 58,8.100%/7,35 = 800(g)

V_ddH2SO4=m_dd/D = 800/1,025= 780,48(ml)

\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{73\cdot36.5\%}{36.5}=0.73\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(1.............2\)

\(0.1.........0.73\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.73}{2}\rightarrow HCldư\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{ZnCl_2}=0.1\cdot136=13.6\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=6.5+73-0.1\cdot2=79.3\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{13.6}{79.3}\cdot100\%=17.15\%\)

\(C\%_{HCl\left(dư\right)}=\dfrac{\left(0.73-0.2\right)\cdot36.5}{79.3}\cdot100\%=25.4\%\)

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