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\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a) Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2
b) \(n_{Fe}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(m_{Cu}=16-11,2=4,8\left(g\right)\)
c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
Chúc bạn học tốt
Gọi x, y lần lượt là sô mol của Fe và Mg
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH:
Fe + H2SO4 ---> FeSO4 + H2 (1)
Mg + H2SO4 ---> MgSO4 + H2 (2)
a. Theo PT(1): \(n_{H_2}=n_{Fe}=x\left(mol\right)\)
Theo PT(2): \(n_{H_2}=n_{Mg}=y\left(mol\right)\)
\(\Rightarrow x+y=0,3\) (*)
Theo đề, ta có: 56x + 24y = 10.4 (**)
Từ (*) và (**), ta có HPT:
\(\left\{{}\begin{matrix}x+y=0,3\\56x+24y=10,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right);m_{Mg}=0,2.24=4,8\left(g\right)\)
b. Ta có: \(n_{hh}=0,1+0,2=0,3\left(mol\right)\)
Theo PT(1,2): \(n_{H_2SO_4}=n_{hh}=0,3\left(mol\right)\)
Đổi 200ml = 0,2 lít
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5M\)
\(n_{H2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
a) Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,5 1 0,5
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O|\)
1 6 2 3
0,1 0,6
b) \(n_{Fe}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{Fe}=0,5.56=28\left(g\right)\)
\(m_{Fe2O3}=44-28=16\left(g\right)\)
0/0Fe = \(\dfrac{28.100}{44}=63,64\)0/0
0/0Fe2O3 = \(\dfrac{16.100}{44}=36,36\)0/0
c) Có : \(m_{Fe2O3}=16\left(g\right)\)
\(n_{Fe2O3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=1+0,6=1,6\left(mol\right)\)
⇒ \(m_{HCl}=1,6.36,5=58,4\left(g\right)\)
\(m_{ddHCl}=\dfrac{58,4.100}{5}=1168\left(g\right)\)
Chúc bạn học tốt
\(a)Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ b)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ n_{Fe}=n_{H_2}=0,1mol\\ m_{Fe}=0,1.56=5,6g\\ m_{Fe_2O_3}=21,6-5,6=16g\\ c)n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\\ n_{H_2SO_4}=0,1+0,1.3=0,4mol\\ C_{M_{H_2SO_4}}=\dfrac{0,4}{0,5}=0,8M\)
a) PTHH: Zn + H2SO4 -> ZnSO4 + H2
nH2= 0,15(mol)
=> nZn=nH2SO4=nZnSO4=nH2=0,15(mol)
b) mZn=0,15.65=9,75(g)
c) CMddH2SO4= 0,15/ 0,05=3(M)
d) mZnSO4= 161. 0,15=24,15(g)
a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
____0,15<--0,3<--------------0,15
=> mFe = 0,15.56 = 8,4 (g)
=> \(\left\{{}\begin{matrix}\%Fe=\dfrac{8,4}{21,2}.100\%=39,62\%\\\%Cu=\dfrac{21,2-8,4}{21,2}.100\%=60,38\%\end{matrix}\right.\)
b) mHCl = 0,3.36,5 = 10,95(g)
=> \(m_{ddHCl}=\dfrac{10,95.100}{3,65}.100\%=300\left(g\right)\)
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,05 0,05
\(m_{Zn}=0,05\cdot65=3,25\left(g\right)\)
\(\Rightarrow m_{Cu}=6,45-3,25=3,2\left(g\right)\)
a, \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
\(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,025\left(mol\right)\Rightarrow m_{C_2H_2}=0,025.26=0,65\left(g\right)\)
b, \(\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,025.22,4}{33,6}.100\%\approx1,67\%\\\%V_{CH_4}\approx98,33\%\end{matrix}\right.\)
m Fe2O3=28.\(\dfrac{75}{100}\)=21g =>n Fe2O3=\(\dfrac{21}{160}\)=0,13125 mol
m CuO=28-21=7 g =>n CuO=\(\dfrac{7}{80}\)=0,0875 mol
Fe2O3+3H2->2Fe+3H2O
0,13125-0,39375-0,2625
CuO+H2-to>Cu+H2O
0,0875-0,0875-0,0875 mol
=>m Fe=0,2625.56=14,7g
=>m Cu=0,0875.64=5,6g
=>n H2=0,39375+0,0875=0,48125mol
ăn r