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![](https://rs.olm.vn/images/avt/0.png?1311)
a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{5,25}.100\%\approx51,43\%\\\%m_{Al_2O_3}\approx48,57\%\end{matrix}\right.\)
b, \(n_{Al_2O_3}=\dfrac{5,25-0,1.27}{102}=0,025\left(mol\right)\)
Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=0,45\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,45.36,5}{29,2\%}=56,25\left(g\right)\)
c, \(n_{H_2SO_4}=\dfrac{1}{2}n_{HCl}=0,225\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,225.98}{19,6\%}=112,5\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
- 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
![](https://rs.olm.vn/images/avt/0.png?1311)
Tiếp bài của creeper nhé:
c. Ta có: \(n_{ZnO}=\dfrac{4,86}{81}=0,06\left(mol\right)\)
Theo PT(1): \(n_{HCl}=2.n_{ZnO}=2.0,06=0,12\left(mol\right)\)
Theo PT(2): \(n_{HCl}=2.n_{Zn}=2.0,1=0,2\left(mol\right)\)
=> \(n_{HCl}=0,12+0,2=0,32\left(mol\right)\)
=> \(m_{HCl}=0,32.36,5=11,68\left(g\right)\)
Ta có: \(C_{\%_{HCl}}=\dfrac{11,68}{m_{dd_{HCl}}}.100\%=12\%\)
=> \(m_{dd_{HCl}}=\dfrac{292}{3}\left(g\right)\)
Theo đề, ta có:
\(D=\dfrac{\dfrac{292}{3}}{V_{dd_{HCl}}}=1,2\)(g/ml)
=> \(V_{dd_{HCl}}=81,1\left(ml\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chọn B
Chỉ có Zn phản ứng với H 2 S O 4 , Cu không phản ứng
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b, Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{4}{15}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{\dfrac{4}{15}.27}{10}.100\%=72\%\\\%m_{Cu}=28\%\end{matrix}\right.\)
c, Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{39,2}{300}.100\%\approx13,067\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Zn+ H2SO4→ ZnSO4+ H2↑
(mol) 0,1 0,1 0,1
a)
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1\left(lít\right)\)
→mZn=n.M=0,1.65= 6,5(g)
→mCu= 10- 6,5= 3,5(g)
=> \(\%m_{Zn}=\dfrac{6,5}{10}.100\%=65\%\)
\(\%m_{Cu}=100\%-65\%=35\%\)
b) \(C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,1}{0,1}=1M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
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