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\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ TheoPT:n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\\ b.2H_2+O_2-^{t^o}\rightarrow2H_2O\\ n_{O_2}=\dfrac{1}{2}n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{O_2}=0,1.32=3,2\left(g\right)\)
a. PTHH: Fe + 2HCl ---> FeCl2 + H2 (1)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo pthh (1): \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
\(\rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, PTHH: 2H2 + O2 --to--> 2H2O (2)
Theo pthh (2): \(n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(\rightarrow m_{O_2}=0,1.32=3,2\left(g\right)\)
\(n_{H_2}=\dfrac{V}{24,79}=\dfrac{11,2}{24,79}\approx0,45\left(mol\right)\)
a) \(PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\)
2 1 2
0,45 0,225 0,45
b) \(m_{O_2}=n.M=0,225.\left(16.2\right)=7,2\left(g\right)\\ V_{O_2}=n.24,79=0,225.24,79=5,57775\left(l\right)\)
c) \(PTHH:2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
2 1 1 1
0,45 0,225 0,225 0,225
\(m_{KMnO_4}=n.M=0,45.\left(39+55+16.4\right)=71,1\left(g\right).\)
a, \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
b, Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{O_2}=0,25.32=8\left(g\right)\)
\(V_{O_2}=0,25.22,4=5,6\left(l\right)\)
c, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,5\left(mol\right)\Rightarrow m_{KMnO_4}=0,5.158=79\left(g\right)\)
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H_2}=n_{Fe}=0.2\left(mol\right)\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(n_{HCl}=2n_{Fe}=0.2\cdot2=0.4\left(mol\right)\)
\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)
\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)
\(n_{Cu}=n_{H_2}=0.2\left(mol\right)\)
\(m_{Cu}=0.2\cdot64=12.8\left(g\right)\)
\(n_{H_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)
\(a.\)
\(2H_2+O_2\underrightarrow{^{^{t^0}}}2H_2O\)
\(0.5.......0.25\)
\(m_{O_2}=0.25\cdot32=8\left(g\right)\)
\(b.\)
\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)
\(0.5............................................0.25\)
\(m_{KMnO_4}=0.5\cdot158=79\left(g\right)\)
- PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
- Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\)
- PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=0,075\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,075.22,4=1,68\left(l\right)\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=8,4\left(l\right)\)
\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ b) n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(kmol)\\ V_{H_2} = 0,2.22,4 = 4,48(m^3)\\ c) 2H_2 + O_2 \xrightarrow{t^o}2H_2O\\ n_{H_2O} = n_{H_2} = 0,2(kmol)\\ m_{H_2O} = 0,2.18 = 3,6(kg)\)
a+b) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)=n_{H_2}\) \(\Rightarrow V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)
c) PTHH: \(H_2+\dfrac{1}{2}O_2\xrightarrow[]{t^o}H_2O\)
Theo PTHH: \(n_{H_2O}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2O}=0,2\cdot18=3,6\left(g\right)\)
nFe = 22,4 : 56 = 0,4 (mol)
pthh : Fe + 2HCl ---> FeCl2+H2
0,4---------------------->0,4(mol)
=> VH2 = 0,4 . 22,4= 8,96 (L)
nO2 = 3,36 : 22,4 = 0,15 (mol)
pthh : 2H2 + O2 ---> 2H2O
LTL :0,4/2 > 0,15/1
=> H2 dư => tính theo O2
theo pt nH2O =2 nO2 = 0,3 (mol)
=> mH2O = 0,3 , 18=5,4 (G)
Câu 1:
PTHH: Fe + 2HCl ===> FeCl2 + H2
a/ nFe = 11,2 / 56 = 0,2 mol
=> nH2 = 0,2 mol
=> VH2(đktc) = 0,2 x 22,4 = 4,48 lít
b/ => nHCl = 0,2 x 2 = 0,4 mol
=> mHCl = 0,4 x 36,5 = 14,6 gam
nFe = 11,2/56 = 0,2 (mol)
PTHH: Fe + 2HCl -> FeCl2 + H2
Mol: 0,2 ---> 0,4 ---> 0,2 ---> 0,2
2H2 + O2 -> (t°) 2H2O
0,2 ---> 0,1
VH2 = 0,2 . 22,4 = 4,48 (l)
mO2 = 0,1 . 32 = 3,2 (g)