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\(n_{FeO}=a\left(mol\right),n_{CuO}=b\left(mol\right)\)
\(m_{hh}=72a+80b=19.2\left(g\right)\left(1\right)\)
\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(n_{H_2SO_4}=a+b=0.25\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.15\)
\(m_{FeO}=0.1\cdot72=7.2\left(g\right)\)
\(m_{CuO}=12\left(g\right)\)
\(C_{M_{FeSO_4}}=\dfrac{0.1}{0.25}=0.4\left(M\right)\)
\(C_{M_{CuSO_4}}=\dfrac{0.15}{0.25}=0.6\left(M\right)\)
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) \(C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,1}{0,1}=1M\)
a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,15.65=9,75\left(g\right)\)
\(\Rightarrow m_{Cu}=m_{hh}-m_{Zn}=21-9,75=11,25\left(g\right)\)
Bạn tham khảo nhé!
Câu 1:a) nH2= 4,48/22,4=0,2(mol)
PTHH: Zn + 2 HCl -> ZnCl2 + H2
0,2_________0,4______0,2______0,2(mol)
nZn= nH2= 0,2(mol) => mZn= 0,2.65=13(g)
\(\%mZn=\frac{13}{21,1}.100\approx61,611\%\\ \Rightarrow\%mZnO\approx100\%-61,611\%\approx38,389\%\)
b) => mZnO= mhhA- mZn= 21,1-13=8,1(g)
=> nZnO= 8,1/81= 0,1(mol)
PTHH: ZnO + 2 HCl -> ZnCl2 + H2O
0,1_________0,2_____0,1(mol)
mZnCl2= 0,2.136+0,1.136=40,8(g)
mddsau= mhhA++mddHCl-mH2= 21,1+200 - 0,2.2= 220,7(g)
=> C%ddZnCl2= (40,8/220,7).100\(\approx18,487\%\)
PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
a________a (mol)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
b________3b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}80a+160b=16\\a+3b=0,25\cdot1=0,25\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,1\cdot80}{16}\cdot100\%=50\%\\\%m_{Fe_2O_3}=50\%\end{matrix}\right.\)
Gọi n CuO = a ( mol )
n Fe2O3 = b ( mol )
Có : n H2SO4 = 0,25 ( mol )
PTHH
CuO + H2SO4 ===> CuSO4 + H2O
a-----------a
Fe2O3 + 3H2SO4 ===> Fe2(SO4)3 + 3H2O
b-----------3b
Ta có hpt
\(\left\{{}\begin{matrix}80a+160b=16\\a+3b=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)
=> m CuO = 8 ( g ) , m Fe2O3 = 8 ( g )
=> %m CuO = %m Fe2O3 = 50 %
Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
a. PTHH:
Zn + H2SO4 ---> ZnSO4 + H2 (1)
MgO + H2SO4 ---> MgSO4 + H2O (2)
b. Theo PT(1): \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)
=> \(m_{Zn}=0,5.65=32,5\left(g\right)\)
(Sai đề nhé.)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
0,5 0,5
b)\(m_{Zn}=0,5\cdot65=32,5\left(g\right)\)
\(m_{ZnO}=\) ko tính đc do lỗi đề
a)
\(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
_____0,4<---0,4<--------0,4<----0,4
=> mZn = 0,4.65 = 26 (g)
=> \(\left\{{}\begin{matrix}\%Zn=\dfrac{26}{51,6}.100\%=50,388\%\\\%Cu=\dfrac{51,6-26}{51,6}.100\%=49,612\text{%}\end{matrix}\right.\)
b)
mZnSO4 = 0,4.161 = 64,4 (g)
c)
\(V_{ddH_2SO_4}=\dfrac{0,4}{2}=0,2\left(l\right)\)
PTHH:
Zn + H2SO4 ---> ZnSO4 + H2 (1)
2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2 (2)
Ta có: \(n_{H_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
Gọi x, y lần lượt là số mol của Zn và Al
a. Theo PT(1): \(n_{H_2}=n_{Zn}=x\left(mol\right)\)
Theo PT(2): \(n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}y\left(mol\right)\)
=> \(x+\dfrac{3}{2}y=0,8\) (*)
Theo đề, ta có: 65x + 27y = 3,79 (**)
Từ (*) và (**), ta có HPT:
\(\left\{{}\begin{matrix}x+\dfrac{3}{2}y=0,8\\65x+27y=3,79\end{matrix}\right.\)
(Ra số âm, bn xem lại đề nhé.)
\(600l\rightarrow600ml\\ n_{Zn}=x;n_{CuO}=y\\ PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ m_{ddH_2SO_4}=1,1.600=660\left(ml\right)=0,66\left(l\right)\\ hpt:\left\{{}\begin{matrix}65x+80y=21\\x+y=0,66.0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,36\\y=-0,03\end{matrix}\right.\left(KTM\right)\)
Xem lại đề giúp mình
.