a. PTHH:

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

Cu + H₂SO₄ ---x--->

b. Theo PT: \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.\dfrac{6,72}{22,4}=0,2\left(mol\right)\)

=> \(m_{Al}=0,2.27=5,4\left(g\right)\)

=> \(m_{Cu}=10-5,4=4,6\left(g\right)\)

c. \(\%_{m_{Al}}=\dfrac{5,4}{10}.100\%=54\%\)

\(\%_{m_{Cu}}=100\%-54\%=46\%\)

d. Theo PT: \(n_{H_2SO_4}=n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{29,4}{m_{dd_{H_2SO_4}}}.100\%=20\%\)

=> \(m_{dd_{H_2SO_4}}=147\left(g\right)\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{Al}=0,2(mol)\Rightarrow m_{Al}=0,2.27=5,4(g)\\ \Rightarrow m_{Cu}=10-5,4=4,6(g)\\ \Rightarrow \%_{Al}=\dfrac{5,4}{10}.100\%=54\%\\ \Rightarrow \%_{Cu}=100\%-54\%=46\%\\ n_{H_2SO_4}=0,3(mol)\Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,3.98}{20\%}=147(g)\)

a, Al + H2SO4 -> Al2(SO4)3 + H2
Fe + H2SO4 -> FeSO4 + H2
b, Gọi số mol của Al là x ( mol ) -> mAl = 27x (g)
Gọi số mol của Fe là y (mol) -> mFe = 56y (g)

=> 27x + 56y = 10 ( I )
nH2 = 6,72/ 22,4 = 0,3 mol
2 Al + 3H2SO4 -> Al2(SO4)3 + 3H2 (1)
2 : 3 : 1 : 3
x : 1,5x : 0,5x : 1,5x

Fe + H2SO4 -> FeSO4 + H2 (2)
1 : 1 : 1 : 1
y : y : y : y
Từ (1) và (2) => 1,5x + y = 0,3 ( II )
Giải I và II => x = 34/285; y=23/190
mAl = 27. 34/285 = 3,22 g
mFe = 10 - 3,22 = 6,78g
 

c, nH2SO4 = 1,5x + y = 0,3
mH2SO4 = 0,3.98 = 29,4 g
mddH2SO4 = (29,4/20): 100 = 147 g





 

\(A.Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ B.n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

0,05      0,05           0,05           0,05

\(\%m_{Mg}=\dfrac{0,05.24}{6,4}\cdot100=18,75\%\\ \%m_{Cu}=100-18,75=81,25\%\\ C.m_{ddH_2SO_4}=\dfrac{0,05.98}{20}\cdot100=24,5g\)

a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

Theo Pt : \(n_{Mg}=n_{H2SO4}=n_{MgSO4}=n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

b)  \(\%m_{Mg}=\dfrac{0,05.24}{6,4}.100\%=18,75\%\)

\(\%m_{Cu}=100\%-18,75\%=81,25\%\)

c) \(m_{H2SO4}=0,05.98=4,9\left(g\right)\)

\(\Rightarrow m_{ddH2SO4}=\dfrac{4.100\%}{20\%}=20\left(g\right)\)

 Chúc bạn học tốt

\(a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b.n_{H_2}=\dfrac{4,48}{22,4}-0,2mol\\ n_{Zn}=n_{H_2SO_4}=n_{H_2}=0,2mol\\ \%m_{Zn}=\dfrac{0,2.65}{19,4}\cdot100=67,01\%\\ \%m_{Cu}=100-67,01=32,99\%\\ m_{ddH_2SO_4}=\dfrac{0,2.98}{20}\cdot100=98g\)

giúp minh với các bạn 

a, Cu không tác dụng với dd HCl.

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)

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

\(\Rightarrow m_{Cu}=19,4-13=6,4\left(g\right)\)

c, Ta có: \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{19,4}.100\%\approx67,01\%\\\%m_{Cu}\approx32,99\%\end{matrix}\right.\)

Bạn tham khảo nhé!