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hỗn hợp 100% <=> 20g
1. fe2o3 60% <=> 12g <=> 0,075mol
cuo 40% <=> 8g <=> 0,1 mol
Ta có:
Fe2O3 + 3H2 --> 2Fe + 3H2O
mol: 0,075 0,225 0,15
CuO + H2 → Cu + H20
mol: 0.1 0.1 0.1
mFe= 0,15x56=8,4g. mCu=0,1x64= 6,4g
nH2= 0,225+0,1=0,325mol ==> V H2 = 0,325x 22,4 = 7,28 lít
\(m_{Fe_2O_3}=16\cdot75\%=12\left(g\right)\)
\(n_{Fe_2O_3}=\dfrac{12}{160}=0.075\left(mol\right)\)
\(n_{CuO}=16\cdot25\%=4\left(g\right)\)
\(n_{CuO}=\dfrac{4}{80}=0.05\left(mol\right)\)
\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(n_{H_2}=3\cdot0.075+0.05=0.275\left(mol\right)\)
a,\(m_{Fe_2O_3}=16.75\%=12\left(g\right)\Rightarrow n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\)
\(m_{CuO}=16-12=4\left(g\right)\Rightarrow n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
PTHH: Fe2O3 + 3H2 → 2Fe + 3H2O
Mol: 0,075 0,225 0,15
PTHH: CuO + H2 → Cu + H2O
Mol: 0,05 0,05 0,05
\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right);m_{Cu}=0,05.64=3,2\left(g\right)\)
b,\(n_{H_2}=0,225+0,05=0,275\left(mol\right)\)
a, mFe2O3 = 32 . 75% = 24 (g)
nFe2O3 = 24/160 = 0,15 (mol)
mCuO = 32 - 24 = 8 (g)
nCuO = 8/80 = 0,1 (mol)
PTHH:
Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O
0,15 ---> 0,45 ---> 0,3
CuO + H2 -> (t°) Cu + H2O
0,1 ---> 0,1 ---> 0,1
mFe = 0,3 . 56 = 16,8 (g)
mCu = 64 . 0,1 = 6,4 (g)
b, nH2 = 0,1 + 0,45 = 0,55 (mol)
VH2 = 0,55 . 22,4 = 12,32 (l)
c, PTHH:
2Al + 6HCl -> 2AlCl3 + 3H2
11/30 <--- 1,1 <--- 11/30 <--- 0,55
mAl = 11/30 . 27 = 9,9 (g)
mHCl = 1,1 . 36,5 = 40,15 (g)
\(a) n_{Fe_2O_3} = \dfrac{16.75\%}{160} = 0,075(mol)\\ n_{CuO} = \dfrac{16.25\%}{80} = 0,05(mol)\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Fe} = 2n_{Fe_2O_3} = 0,15(mol) \Rightarrow m_{Fe} = 0,15.56 = 8,4(gam)\\ n_{Cu} = n_{CuO} = 0,05(mol) \Rightarrow m_{Cu} = 0,05.64 = 3,2(gam)\\ b) n_{H_2} = 3n_{Fe_2O_3} + n_{CuO} = 0,075.3 + 0,05 = 0,275(mol)\\ V_{H_2} = 0,275.22,4 = 6,16(lít)\)
, PTHH :
Fe2O3 + 3H2 -> 2Fe + 3H2O
0,075......0,225.........0,15 (mol)
CuO + H2 -> Cu + H2O
0,1.......0,1......0,1 (mol)
mFe2O3 = 20.60%=12(g) => nFe2O3 = 0,075(mol)
mCuO = 8 ( g) => nCuO = 0,1(mol)
mFe = 0,15 .56=8,4(g)
mCu = 0,1 .64=6,4(g)
=>bạn llàm như ý đướiưới để tính % nhé
Ta có:
\n\n\\(\\%m_{Fe2O3}=60\\%.20=12\\left(g\\right)\\Rightarrow n_{Fe2O3}=\\frac{12}{160}=0,075\\left(mol\\right)\\)
\n\n\\(\\%m_{CuO}=40\\%.20=8\\left(g\\right)\\Rightarrow n_{CuO}=\\frac{8}{80}=0,1\\left(mol\\right)\\)
\n\n\\(Fe_2O_3+3H_2\\rightarrow2Fe+3H_2O\\)
\n\n_0,075__________0,15_____
\n\n\\(CuO+H_2\\rightarrow Cu+H_2O\\)
\n\n_0,1___________0,1_____
\n\n\\(\\left\\{{}\\begin{matrix}m_{Fe}=0,15.56=8,4\\left(g\\right)\\\\m_{Cu}=0,1.64=6,4\\left(g\\right)\\end{matrix}\\right.\\)
\n\n\\(\\Rightarrow\\left\\{{}\\begin{matrix}\\%m_{Fe}=\\frac{8,4}{8,4+6,4}.100\\%=56,76\\%\\\\\\%m_{Cu}=100\\%-56,76\\%=43,24\\%\\end{matrix}\\right.\\)
\n
PTHH
Fe2O3 + 3H2 -> 2Fe + 3H2O
0.075 0.225 mol
CuO + H2 -> Cu + H2O
0.1 0.1 mol
a) mFe2O3 = 20*60%=12g
mCuO = 20 - 12 = 8g
b) nFe2O3 = 12/160 = 0.075mol
nCuO = 8/80 = 0.1 mol
VH2 = (0.225+0.1)*22.4 = 7.28 lit