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Fe+3Cl\(\underrightarrow{t^o}\)FeCl3
mFe+mCl=mFeCl3
BTKL: mFe+mCl=mFeCl3
11,2 +21,3=mFeCl3
=>mFeCl3=32,5(gam)
VCl(đkt)=24.0,9=21,6 lít
\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\\
pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
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, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT: \(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)
\(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
\(n_{Al}=\dfrac{9,45}{27}=0,35\left(mol\right)\\ a,PTHH:2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ b,n_{Cl_2}=\dfrac{3}{2}.0,35=0,525\left(mol\right)\\ V_{Cl_2\left(đkc\right)}=0,525.24,79=13,01475\left(l\right)\\ c,n_{AlCl_3}=n_{Al}=0,35\left(mol\right)\\ m_{AlCl_3}=0,35.133,5=46,725\left(g\right)\)
1.\(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,3 0,1 ( mol )
\(m_{Fe}=0,3.56=16,8g\)
2.\(n_{Cu}=\dfrac{3,2}{64}=0,05mol\)
\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)
0,05 0,05 ( mol )
\(m_{CuO}=0,05.80=4g\)
3.\(n_{Na}=\dfrac{4,6}{23}=0,2mol\)
\(4Na+O_2\rightarrow\left(t^o\right)2Na_2O\)
0,2 0,05 ( mol )
\(V_{O_2}=0,05.24,79=1,2395l\)
4.\(n_{Cu}=\dfrac{1,6}{64}=0,025mol\)
\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)
0,025 0,0125 ( mol )
\(V_{O_2}=0,0125.24,79=0,309875l\)
Bài 2: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PTHH: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)
Theo PTHH: \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=127\cdot0,1=12,7\left(g\right)\)
\(n_{HCl}=0,3.1=0,3\left(mol\right)\\
pthh:Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15
\(m_{Fe}=0,15.56=8,4\left(g\right)\\
V_{H_2}=0,15.22,4=3,36\left(l\right)\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{Cl_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\\ a,V_{Cl_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{FeCl_3}=162,5.0,2=32,5\left(g\right)\)
a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PT: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)
Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\)
\(\Rightarrow V_{Cl_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PT: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)
Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\Rightarrow V_{Cl_2}=0,3.24,79=7,437\left(l\right)\)
\(n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)