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a) \(n_{Fe}=\dfrac{1,12}{56}=0,02\left(mol\right)\)
PTHH: Fe + 2HCl -->FeCl2 + H2
_____0,02->0,04--->0,02--->0,02
=> VH2 = 0,02.22,4 = 0,448(l)
b) mFeCl2 = 0,02.127 = 2,54(g)
c) \(C_{M\left(HCl\right)}=\dfrac{0,04}{0,2}=0,2M\)
Fe + 2HCl → FeCl2 + H2
1 2 1 1
0,02 0,04 0,02 0,02
nFe=\(\dfrac{1,12}{56}\)= 0,02(mol)
a). nH2=\(\dfrac{0,02.1}{1}\)= 0,02(mol)
→VH2= n . 22,4 = 0,02 . 22,4 = 0,448(l)
b). nFeCl2= \(\dfrac{0,02.1}{1}\)= 0,02(mol)
→mFeCl2= n . M = 0,02 . 127 = 2,54(g)
c). 200ml = 0,2l
nHCl= \(\dfrac{0,02.2}{1}\)=0,04(mol)
→CM= \(\dfrac{n}{V}\)= \(\dfrac{0,04}{0,2}\)= 0,2M
Ta có: \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
______0,5____0,5_____0,5_____0,5 (mol)
b, mH2SO4 = 0,5.98 = 49 (g)
c, mFeSO4 = 0,5.152 = 76 (g)
d, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
____0,5__0,5 (mol)
⇒ mCu = 0,5.64 = 32 (g)
Bạn tham khảo nhé!
Gọi \(n_{Fe}=x\left(mol\right)\)\(;n_{Zn}=y\left(mol\right)\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)
Ta có: \(\left\{{}\begin{matrix}56x+65y=18,6\\2x+2y=2n_{H_2}=0,6\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\%m_{Fe}=\dfrac{0,1\cdot56}{18,6}\cdot100\%=30,11\%\)
\(\%m_{Zn}=100\%-30,11\%=69,89\%\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4
\(n_{HCl}=0,2+0,4=0,6mol\)
\(C_M=\dfrac{n}{V}=\dfrac{0,6}{0,2}=3M\)
\(Fe+H_2SO_4 \to FeSO_4+H_2\\ n_{H_2}=0,15(mol)\\ a/\\ n_{Fe}=n_{H_2}=0,15(mol)\\ m_{Fe}=0,15.56=8,4(g)\\ b/\\ n_{H_2SO_4}=n_{H_2}=0,15(mol)\\ CM_{H_2SO_4}=\dfrac{0,15}{2}=0,75M c/\\ n_{FeSO_4}=n_{H_2}=0,15(mol)\\ CM_{FeSO_4}=\dfrac{0,15}{0,2}=0,75M\\\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,1 0,1 0,1 0,1
\(m_{Fe}=0,1\cdot56=5,6\left(g\right)\)
\(m_{FeSO_4}=0,1\cdot152=15,2\left(g\right)\)
\(C_M=\dfrac{0,1}{0,2}=0,5M\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a. PTHH: Fe + H2SO4 ---> FeSO4 + H2
Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)
=> \(m_{Fe}=0,1.56=5,6\left(g\right)\)
b. Theo PT: \(n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\)
=> \(m_{FeSO_4}=0,1.152=15,2\left(g\right)\)
c. Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\)
Đổi 200ml = 0,2 lít
=> \(C_{M_{H_2SO_4}}=\dfrac{0,1}{0,2}=0,5M\)