\(n_{H_2SO_4}=0,25.2=0,5\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Al_2O_3}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{matrix}\right.\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

 x----------> 3x --------> x 

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

y --------> y -------->  y

Có hệ phương trình

\(\left\{{}\begin{matrix}102x+80y=26,2\\3x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\%_{m_{Al_2O_3}}=\dfrac{102.0,1.100}{26,2}=38,93\%\)

\(\%_{m_{CuO}}=\dfrac{80.0,2.100}{26,2}=61,07\%\)

\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{x}{0,25}=\dfrac{0,1}{0,25}=0,4M\)

\(CM_{CuSO_4}=\dfrac{y}{0,25}=\dfrac{0,2}{0,25}=0,8M\)

\(a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ b,n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{Al}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ c,n_{H_2SO_4}=0,3(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,3.98}{9,8\%}=300(g)\\ d,n_{Al_2(SO_4)_3}=0,1(mol)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{5,4+300-0,3.2}.100\%=11,22\%\)

\(a)n_{H_2}=\dfrac{7,437}{24,79}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2        0,3               0,1                  0,3

\(\%m_{Al}=\dfrac{0,2.27}{15,6}\cdot100=34,62\%\\ \%m_{Al_2O_3}=100-34,62=65,38\%\\ b)n_{Al_2O_3}=\dfrac{\left(15,6-0,2.27\right)}{102}=0,1mol\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

0,1        0,15           0,05                 0,15

\(m_{ddH_2SO_4}=\dfrac{\left(0,3+0,15\right).98}{9,8}\cdot100=450g\)

c) \(C_{\%Al_2\left(SO_4\right)_3}=\dfrac{\left(0,1+0,05\right)342}{15,6+450-0,3.2}\cdot100=11\%\)

\(n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

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

\(0.05.......0.05......0.05...........0.05\)

\(m_{Zn}=0.05\cdot65=3.25\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{0.05}{0.2}=0.25\left(M\right)\)

\(m_{ZnSO_4}=0.05\cdot161=8.05\left(g\right)\)

\(CuO+H_2SO_{4\left(24,5\%\right)}\rightarrow CuSO_4+H_2O\)

\(Cu+2H_2SO_{4đ}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)

\(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(\Rightarrow n_{Cu}=0,05\left(mol\right)\)

\(\Rightarrow m_{CuO}=10-64.0,05=6,8\left(g\right)\)

\(\Rightarrow n_{CuO}=0,085\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(24,5\%\right)}=0,085\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(24,5\%\right)}=8,33\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4\left(24,5\%\right)}=34\left(g\right)\)

a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,5.65=32,5\left(g\right)\)

\(\Rightarrow m_{CuO}=72,5-32,5=40\left(g\right)\)

c, Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Zn}+n_{CuO}=1\left(mol\right)\)

\(\Rightarrow b=C_{M_{H_2SO_4}}=\dfrac{1}{2,5}=0,4M\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnSO_4}=n_{Zn}=0,5\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnSO_4}}=\dfrac{0,5}{2,5}=0,2M\\C_{M_{CuSO_4}}=\dfrac{0,5}{2,5}=0,2M\end{matrix}\right.\)

Bạn tham khảo nhé!

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\left(I\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\left(II\right)\)

b, Theo PTHH(1) : \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{Zn}=32,5\left(g\right)\)

\(\Rightarrow m_{CuO}=m_{hh}-m_{Zn}=40\left(g\right)\)

\(\Rightarrow n_{CuO}=\dfrac{m}{M}=0,5\left(mol\right)\)

c, Theo PTHH (1) và (2) : \(n_{H2SO4}=n_{CuO}+n_{Zn}=1\left(mol\right)\)

\(\Rightarrow C_{MH2SO4}=b=\dfrac{n}{V}=\dfrac{1}{2,5}=0,4M\)

d, ( Chắc là thể tích coi như không đổi )

Thấy sau phản ứng thu được A gồm \(0,5molZnSO_4,0,5molCuSO_4\)

\(\Rightarrow C_{MCuSO4}=C_{MZnSO4}=\dfrac{n}{V}=\dfrac{0,5}{2,5}=0,2M\)

Vậy ...

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{Fe} = n_{H_2} = \dfrac{3360}{1000.22,4} = 0,15(mol)$
$n_{SO_2} = \dfrac{4480}{22,4.1000} = 0,2(mol)$
Bảo toàn electron : 

$3n_{Fe} + 2n_{Cu} = 2n_{SO_2}$

Suy ra: $0,15.3 + 2n_{Cu} = 0,2.2 \Rightarrow n_{Cu} = -0,025<0$

(Sai đề)