Chọn B

Chỉ có Zn phản ứng với H 2 S O 4 , Cu không phản ứng

Chọn A

- Cu không tác dụng được với dd H₂SO₄ loãng.

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{Zn}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Zn}=\dfrac{0,1.65}{10,5}.100\approx61,905\%\\ \Rightarrow\%m_{Cu}\approx38,095\%\)

anh giúp em bài này  với https://hoc24.vn/cau-hoi/giup-minh-voi-trong-tam-giai-thich-ki-cai-de-nha-cam-on.2017646398420

$Zn + H_2SO_4 \to ZnSO_4 + H_2$

Theo PTHH : 

$n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$\%m_{Zn} = \dfrac{0,1.65}{10,5}.100\% = 61,9\%$

Chọn đáp án A

thanks

a. PTHH:

\(Zn+H_2SO_4--->ZnSO_4+H_2\)

\(Cu+H_2SO_4--\times-->\)

b. Theo PT: \(n_{Zn}=n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow m_{_{ }Zn}=0,1.65=6,5\left(g\right)\)

\(\Rightarrow\%_{m_{Zn}}=\dfrac{6,5}{10,5}.100\%=61,9\%\)

\(\%_{m_{Cu}}=100\%-61,9\%=31,8\%\)

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

Cu không phản ứng H₂SO₄ loãng nhé

\(n_{H_2}= \dfrac{2,24}{22,4}= 0,1 mol\)

Theo PTHH:

\(n_{Zn}=n_{H_2}= 0,1 mol\)

\(\Rightarrow m_{Zn}= 0,1 . 65= 6,5 g\)

\(\Rightarrow\)%m_Zn=\(\dfrac{6,5}{10,5} . 100\)%~ 61,9%

\(\Rightarrow\)%m_Cu= 100% - 61,9%=38,1 %

\(a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b.n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{Zn}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Zn}=\dfrac{0,1.65}{10,5}.100=61,9\%\\ \%m_{Cu}=100-61,9=38,1\%\)

10,5 tính như nao ạ?

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Zn + H₂SO₄ --> ZnSO₄ + H₂

_____0,1<-------------------------0,1

=> m_Zn = 0,1.65 = 6,5 (g)
\(\left\{{}\begin{matrix}\%Zn=\dfrac{6,5}{10,5}.100\%=61,9\%\\\%Cu=\dfrac{10,5-6,5}{10,5}.100\%=38,1\%\end{matrix}\right.\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

tl1..........1................1.............1(mol)

br  x.......x................x.............x(mol)

\(Cu+H_2SO_4\rightarrow CuSO_4+H_2\)

tl1............1...............1...........1(mol)

Br y...........y...............y...........y(mol)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Taco hệ pt

\(\left\{{}\begin{matrix}x+y=0,05\\24x+64y=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,045\\y=0,095\end{matrix}\right.\)

\(\Rightarrow\%m_{Mg}=0,045.24:5.100\%=21,6\%\)

\(\Rightarrow\%m_{Cu}=100\%-21,6\%=78,4\%\)

           Zn+       H₂SO₄→           ZnSO₄+    H₂↑

(mol)  0,1           0,1                                    0,1

a)

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1\left(lít\right)\)

→m_Zn=n.M=0,1.65= 6,5(g)

→m_Cu= 10- 6,5= 3,5(g)

=> \(\%m_{Zn}=\dfrac{6,5}{10}.100\%=65\%\)

      \(\%m_{Cu}=100\%-65\%=35\%\)

b) \(C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,1}{0,1}=1M\)