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\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\\
pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,25 0,25 0,25
=> \(V_{H_2}=0,25.22,4=5,6\left(l\right)\)
\(m_{FeSO_4}=152.0,25=38\left(g\right)\)
\(pthh:CuO+H_2\underrightarrow{t^o}H_2O+Cu\)
0,25 0,25 0,25
=> \(m_{Cu}=0,25.64=16\left(g\right)\)
nFe = 14/56 =0,25 mol
PTHH : Fe + H2SO4 => FeSO4 + H2 (1)
Theo pt(1) : nH2 = nFe = 0,25 mol
VO2 = 0,25 x 22,4 = 5,6 l
Theo pt(1): nFeSO4 = nFe = 0,25 mol
mFeSO4= 0,25 x 152 = 38 g
PTHH : H2 + CuO => Cu + H2O(2)
theo pt (2) => nH2 = nCu = 0,25 mol
mCu = 0,25 x 64 = 16 g
\(a\)) \(PTHH:Zn+2HCl\underrightarrow{t^o}ZnCl_2+H_2\)
0,25 0,5 0,25 0,25
b) nZn=\(\dfrac{m}{M}=\dfrac{16,25}{65}=0,25\left(mol\right)\)
\(V_{H_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)
c) \(m_{HCl}=n.M=0,5.36,5=18,25\left(g\right)\)
a) Zn + 2HCl --> ZnCl2 + H2
b) \(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,02-->0,04---------->0,02
=> VH2 = 0,02.22,4 = 0,448 (l)
c) mHCl = 0,04.36,5 = 1,46 (g)
a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT: \(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)
\(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
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): \(n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(\rightarrow m_{O_2}=0,1.32=3,2\left(g\right)\)
\(a) Mg + 2HCl \to MgCl_2 + H_2\\ b) n_{MgCl_2} = n_{Mg} = \dfrac{0,24}{24} = 0,01(mol)\\ m_{MgCl_2} = 0,01.95 = 0,95(gam)\\ c) n_{H_2} = n_{Mg} = 0,01(mol) \Rightarrow V_{H_2} = 0,01.22,4 = 0,224(lít)\)
\(n_{HCl}=0,3.1=0,3\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15
\(m_{Fe}=0,15.56=8,4\left(g\right)\\ V_{H_2}=0,15.22,4=3,36\left(l\right)\)