\(600l\rightarrow600ml\\ n_{Zn}=x;n_{CuO}=y\\ PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ m_{ddH_2SO_4}=1,1.600=660\left(ml\right)=0,66\left(l\right)\\ hpt:\left\{{}\begin{matrix}65x+80y=21\\x+y=0,66.0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,36\\y=-0,03\end{matrix}\right.\left(KTM\right)\)

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ko chắc

a. Ta có: nH₂SO₄ = 0,6 mol

Zn + H₂SO₄ ➝ ZnSO₄ + H₂

CuO + H₂SO₄ ➝ CuSO₄ + H₂O

Gọi số mol của Zn và CuO lần lượt là x, y ta có hệ:

\(\left\{{}\begin{matrix}65x+80y=42\\x+y=0,6\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Zn}=26g\\m_{CuO}=16g\end{matrix}\right.\)

b. m_muối = mZnSO₄ +mCuSO₄ = 96,4 g

m_dd = 1,1.600 = 660 g

=> C% = \(\frac{96,4}{660}100\%\) ≃ 14,6 %

a) Gọi số mol Al, Zn là a, b (mol)

=> 27a + 65b = 11,9 (1)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

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

             a----->1,5a----------------->1,5a

            Zn + H₂SO₄ --> ZnSO₄ + H₂

             b------>b------------------>b

=> 1,5a + b = 0,4 (2)

(1)(2) => a = 0,2 (mol); b = 0,1 (mol)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{11,9}.100\%=45,378\%\\\%m_{Zn}=\dfrac{0,1.65}{11,9}.100\%=54,622\%\end{matrix}\right.\)

b) n_H2SO4 = 1,5a + b = 0,4 (mol)

=> m_H2SO4 = 0,4.98 = 39,2 (g)

=> \(C\%_{dd.H_2SO_4}=\dfrac{39,2}{150}.100\%=26,133\%\)

\(PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\\ CuO+H_2SO_4\to CuSO_4+H_2O\\ \Rightarrow n_{Zn}=n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ \Rightarrow m_{Zn}=0,2.65=13(g)\\ \Rightarrow \%_{Zn}=\dfrac{13}{21}.100\%=61,9\%\\ \Rightarrow \%_{CuO}=100\%-61,9\%=38,1\%\\ \Rightarrow n_{CuO}=\dfrac{21-13}{80}=0,1(mol)\\ \Rightarrow \Sigma n_{H_2SO_4}=0,1+0,2=0,3(mol)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{0,3}{0,5}=0,6(l)\)

\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

a) Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)

            1           1                  1              1

          0,1         0,1                           

           \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2|\)

              1            1          1             1

             0,2         0,2                       0,2             

b) \(n_{Zn}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{Zn}=0,2.65=13\left(g\right)\)

\(m_{CuO}=21-13=8\left(g\right)\)

⁰/₀CuO = \(\dfrac{8.100}{21}=38,1\)⁰/₀

⁰/₀Zn = \(\dfrac{13.100}{21}=61,9\)⁰/₀

c) Có : \(m_{CuO}=8\left(g\right)\)

\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(n_{H2SO4\left(tổng\right)}=0,1+0,2=0,3\left(mol\right)\)

\(V_{ddH2SO4}=\dfrac{0,3}{0,5}=0,6\left(l\right)\)

 Chúc bạn học tốt

a) Sửa đề: dd H₂SO₄ 9,8%

Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2}=0,35\cdot2=0,7\left(g\right)\)

Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{H_2}=0,35\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,35\cdot98}{9,8\%}=350\left(g\right)\)

\(\Rightarrow m_{dd}=m_{KL}+m_{H_2SO_4}-m_{H_2}=361,6\left(g\right)\)

b) Tương tự câu a

\(n_{H_2}=\dfrac{3,36}{22,4}0,15(mol)\\ a,PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ b,n_{Fe}=n_{H_2}=0,15(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,15.56}{14,8}.100\%=56,76\%\\ \Rightarrow \%_{Cu}=100\%-56,76\%=43,24\%\\ c,n_{H_2SO_4}=0,15(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,15.98}{20\%}=73,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{73,5}{1,4}=52,5(l)\)