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\(n_{Al}=\dfrac{10,8}{27}=0,4(mol)\\ a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,6(mol);n_{Al_2(SO_4)_3}=0,2(mol)\\ b,V_{H_2}=0,6.22,4=13,44(l)\\ c,m_{dd_{H_2SO_4}}=\dfrac{0,6.98}{9,8\%}=600(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,2.342}{10,8+600-0,6.2}.100\%=11,22\%\)
\(a.n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \rightarrow\left\{{}\begin{matrix}27a+24b=5,1\\1,5a+b=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,1}.100\approx52,941\%\\\%m_{Mg}\approx47,059\%\end{matrix}\right.\)
\(b.m_{ddH_2SO_4}=\dfrac{0,25.98.100}{9,8}=250\left(g\right)\\ m_{ddsau}=m_{Al,Mg}+m_{ddH_2SO_4}-m_{H_2}=5,1+250-0,25.2=254,6\left(g\right)\\ C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342}{254,6}.100\approx6,716\%\\ C\%_{ddMgSO_4}=\dfrac{0,1.120}{254,6}.100\approx4,713\%\)
\(a/\\3Al+2H_2SO_4 \to Al_2(SO_4)_3+3H_2\\ Zn+H_2SO_4 \to ZnSO_4+H_2\\ n_{H_2SO_4}=\frac{400.9.8\%}{98}=0,4(mol)\\ n_{Al}=a(mol)\\ n_{Zn}=b(mol) m_{hh}=27a+65b=11,9(1)\\ n_{H_2SO_4}=1,5a+b=0,4(mol)\\ (1)(2)\\ a=0,2; b=0,1\\ b/\\ \%m_{Al}=\frac{0,2.27}{11,9}.100=45,38\%\\ \%m_{Zn}=54,62\% \)
a)
$n_{Al} = 0,3(mol)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
Theo PTHH :
$n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,45.98}{12,25\%} = 360(gam)$
b)
$n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$
c)
$n_{Al_2(SO_4)_3} = 0,15(mol)$
$m_{dd\ sau\ pư} = 8,1 + 360 - 0,45.2 = 367,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{367,2}.100\% = 14\%$
1)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
____0,1----->0,15
=> mH2SO4 = 0,15.98 = 14,7(g)
=> \(C\%=\dfrac{14,7}{250}.100\%=5,88\%\)
2)
\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)
PTHH: Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O
_______0,2------------------------------>0,2
=> VCO2 = 0,2.22,4 = 4,48(l)
3)
\(n_A=\dfrac{18,4}{M_A}\left(mol\right)\)
PTHH: 2A + Cl2 --to--> 2ACl
____\(\dfrac{18,4}{M_A}\)---------->\(\dfrac{18,4}{M_A}\)
=> \(\dfrac{18,4}{M_A}\left(M_A+35,5\right)=46,8=>M_A=23\left(Na\right)\)
4)
nHCl = 0,2.3 = 0,6(mol)
PTHH: M + 2HCl --> MCl2 + H2
____0,3<-----0,6
=> \(M_M=\dfrac{7,2}{0,3}=24\left(Mg\right)\)
Ta có: \(n_{Fe_2O_3}=\dfrac{9,6}{160}=0,06\left(mol\right)\)
a. PTHH: Fe2O3 + 3H2SO4 ---> Fe2(SO4)3 + 3H2O (1)
Theo PT(1): \(n_{H_2SO_4}=3.n_{Fe_2O_3}=3.0,06=0,18\left(mol\right)\)
=> \(m_{H_2SO_4}=0,18.98=17,64\left(g\right)\)
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{17,64}{m_{dd_{H_2SO_4}}}.100\%=9,8\%\)
=> \(m_{dd_{H_2SO_4}}=180\left(g\right)\)
b. Ta có: \(m_{dd_{Fe_2\left(SO_4\right)_3}}=9,6+180=189,6\left(g\right)\)
Theo PT(1): \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,06\left(mol\right)\)
=> \(m_{Fe_2\left(SO_4\right)_3}=0,06.400=24\left(g\right)\)
=> \(C_{\%_{Fe_2\left(SO_4\right)_3}}=\dfrac{24}{189,6}.100\%=12,66\%\)
c. PTHH: Fe2(SO4)3 + 3BaCl2 ---> 3BaSO4↓ + 2FeCl3 (2)
Theo PT(2): \(n_{BaSO_4}=3.n_{Fe_2\left(SO_4\right)_3}=3.0,06=0,18\left(mol\right)\)
=> \(m_{BaSO_4}=0,18.233=41,94\left(g\right)\)
Theo PT(2): \(n_{BaCl_2}=n_{BaSO_4}=0,18\left(mol\right)\)
=> \(m_{BaCl_2}=0,18.208=37,44\left(g\right)\)
Ta có: \(C_{\%_{BaCl_2}}=\dfrac{37,44}{m_{dd_{BaCl_2}}}.100\%=10,4\%\)
=> \(m_{dd_{BaCl_2}}=360\left(g\right)\)
Tóm tắt
\(V_{H_2\left(đktc\right)}=8,96l\\ C_{\%H_2SO_4}=19,6\%\\ a)m_{Zn}=?\\ m_{ddH_2SO_4}=?\\ b)C_{\%ZnSO_4}=?\)
\(a)n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,4 0,4 0,4 0,4
\(m_{Zn}=0,4.65=26g\\ m_{ddH_2SO_4}=\dfrac{0,4.98}{19,6}\cdot100=200g\\ b)C_{\%ZnSO_4}=\dfrac{0,4.161}{26+200-0,4.2}\cdot100=28,6\%\)
a)
$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{H_2SO_4} = n_{H_2} = n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)$
$V = 0,3.22,4 = 6,72(lít)$
$C_{M_{H_2SO_4}} = \dfrac{0,3}{0,25} = 1,2M$
b)
$n_{CuO} = \dfrac{16}{80} = 0,2(mol)$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$n_{CuO} < n_{H_2}$ nên $H_2$ dư
$n_{Cu} = n_{CuO} = 0,2(mol)$
$m_{Cu} = 0,2.64 = 12,8(gam)$
\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ PT:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,3 0,3 0,3 (mol)
a) V= n. 22,4 = 0,3 . 22,4 = 6,72(l)
\(C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%=\dfrac{0,3.98}{200}.100\%=11,76\%\)
b) PT: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
0,3 0,3
=> mCu=n.M=0,3.64=19,2(g)
\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,15 0 ,05 0,15
a)\(V=0,15\cdot22,4=3,36\left(l\right)\)
b)\(m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,7}{9,8}\cdot100=150\left(g\right)\)
c) \(m_{H_2}=0,15\cdot2=0,3\left(g\right)\)
\(m_{ddsau}=2,7+150-0,3=152,4\left(g\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\)
\(\Rightarrow C\%=\dfrac{17,1}{152,4}\cdot100=11,22\%\)