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PTHH: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2.
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{200}.100\%=14,7\%\)
=> \(m_{H_2SO_4}=29,4\left(g\right)\)
=> \(n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)
Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)
=> \(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)
mH2SO4 = 9,8%.200=19,6(g) -> nH2SO4=0,2(mol)
nAl=0,1(mol)
PTHH: 2Al +3 H2SO4 -> Al2(SO4)3 + 3H2
Ta có: 0,1/2 < 0,2/3 => H2SO4 dư, Al hết, tính heo nAl
nAl2(SO4)3=mAl/2=0,1/2=0,05(mol)
mAl2(SO4)3 (LT)= 0,05.342=17,1(g)
Vì: H=75%
=>mAl2(SO4)3 (TT)=75%. 17,1=12,825(g)
\(\left(a\right)2Al+3H_2O\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \left(b\right)n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{Al}=\dfrac{0,6.2}{3}=0,4mol\\ m_{Al}=0,4.27=10,8g\\ \left(c\right)n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4Al+3O_2\underrightarrow{t^0}2Al_2O_3\\ \Rightarrow\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Al.dư\\ n_{Al_2O_3}=\dfrac{0,15.2}{3}=0,1mol\\ m_{oxit}=m_{Al_2O_3}=0,1.102=10,2g\)
a: \(2Al+3H_2SO_4\rightarrow1Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,4 0,6 0,2 0,6
b: \(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)
=>\(n_{Al}=0.4\left(mol\right)\)
\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)
c: \(4Al+3O_2\rightarrow2Al_2O_3\)
0,4 0,2
\(m_{Al_2O_3}=0.2\left(27\cdot2+16\cdot3\right)=0.2\cdot102=20.4\left(g\right)\)
Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{6}\left(mol\right)\Rightarrow m_{Al}=\dfrac{1}{6}.27=4,5\left(g\right)\)
b, \(n_{H_2SO_4}=n_{H_2}=0,25\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)
c, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{12}\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{12}.342=28,5\left(g\right)\)
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, \Rightarrow n_{CaCl_2}=n_{CaCO_3}=n_{CO_2}=0,1\left(mol\right)\\ \Rightarrow a=m_{CaCO_3}=100.0,1=10\left(g\right)\\b,n_{HCl}=2.n_{CO_2}=2.0,1=0,2\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(l\right)\\ c,m_{CaCl_2}=111.0,1=11,1\left(g\right)\)
\(n_{CO2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Pt : \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O|\)
1 2 1 1 1
0,1 0,2 0,1 0,1
a) \(n_{CaCO3}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{CaCO3}=0,1.100=10\left(g\right)\)
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(V_{HCl}=\dfrac{0,2}{2}=0,1\left(l\right)\)
c) \(n_{CaCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{CaCl2}=0,1.111=11,1\left(g\right)\)
Chúc bạn học tốt
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Mol: 0,03 0,1 0,03 0,05
mAl=0,03.27=0,9 (g)
mHCl = 0,1.36,5 = 3,65 (g)
\(m_{AlCl_3}=0,03.133,5=4,005\left(g\right)\)
Sửa: \(14,7\%\)
\(n_{H_2SO_4}=\dfrac{200.14,7\%}{100\%.98}=0,3(mol)\\ a,PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=0,3(mol)\\ \Rightarrow V_{H_2}=0,3.22,4=6,72(l)\\ b,n_{Al}=\dfrac{2}{3}n_{H_2SO_4}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ c,n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,1(mol)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{200+5,4-0,3.2}.100\%=16,7\%\\ c,m_{Al_2(SO_4)_3}=0,1.342=34,2(g)\)