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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)
a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a_____2a______a_____a (mol)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
b_____3b_______b_____\(\dfrac{3}{2}\)b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}56a+27b=36,1\\a+\dfrac{3}{2}b=\dfrac{21,28}{22,4}=0,95\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=0,3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,5\cdot56=28\left(g\right)\\m_{Al}=8,1\left(g\right)\end{matrix}\right.\)
b+c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{HCl}=2a+3b=1,9\left(mol\right)\\n_{FeCl_2}=0,5\left(mol\right)\\n_{AlCl_3}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{1,9}{0,2}=9,5\left(M\right)\\C_{M_{FeCl_2}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\)
\(n_{Al}=a\left(mol\right)\)
\(n_{Fe}=b\left(mol\right)\)
\(m=27a+56b=19.3\left(g\right)\left(1\right)\)
\(n_{H^+}=0.2\cdot2+0.2\cdot2.25\cdot2=1.3\left(mol\right)\)
\(2Al+6H^+\rightarrow2Al^{3+}+3H_2\)
\(Fe+2H^+\rightarrow Fe^{2+}+H_2\)
\(n_{H^+}=3a+2b=1.3\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.3,b=0.2\)
\(\%Al=\dfrac{0.3\cdot27}{19.3}\cdot100\%=41.96\%\)
\(\%Fe=58.04\%\)
\(b.\)
\(n_{H_2}=\dfrac{1}{2}n_{H^+}=0.65\left(mol\right)\)
Bảo toàn khối lượng :
\(m_{Muối}=19.3+0.4\cdot36.5+0.45\cdot98-0.65\cdot2=76.7\left(g\right)\)
a) PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b) \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
_____0,02<---0,03<---------------------0,03
=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,02.27}{2,16}.100\%=25\%\\\%Cu=100\%-25\%=75\%\end{matrix}\right.\)
c) mH2SO4 = 0,03.98 = 2,94 (g)
=> \(C\%\left(H_2SO_4\right)=\dfrac{2,94}{200}.100\%=1,47\%\)
2Al + 6HCl -> 2AlCl3 + 3H2 (1)
ZnO + 2HCl -> ZnCl2 + H2O (2)
a) nH2= 13,44/22.4=0.6(mol) -> mH2=0,6.2=1,2(g)
Theo PTHH: nAl = 2/3 nH2 = 2/3 . 0,6= 0,4(mol) -> mAl = 0,4 . 27=10,8(g)
-> mZnO = 27-10,8= 16,2(g)
b) nZnO = 16,2/81=0,2(mol)
Theo PTHH (2): nHCl = 2nZnO=2.0,2=0,4(mol)
Theo PTHH (1) : nHCl=2nH2=2.0,6=1,2(mol)
-> \(\Sigma\)nHCl = 0,4+1,2=1,6(mol)
-> mHCl = 1,6.36,5= 58,4(g)
-> mddHCl = 58,4.100/29,2= 200(g)
c) Theo PTHH (1): nAlCl3 = 2/3 nH2 = 2/3 . 0,6=0,4(mol) -> mAlCl3=0,4.133,5=53,4(g)
mdd sau phản ứng= mA + mddHCl - mH2 =27+200-1,2 =225,8(g)
-> C% AlCl3 = 53,4.100%/225,8 = 20,88%
Theo PTHH (2) nZnCl2 =nZnO= 0,2(mol)-> mZnCl2=0,2.136=27,2(g)
-> C% ZnCl2= 27,2.100%/255,8=10,63%
Bài 4 :
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
b) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)0/0
d) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)
\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)
\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)0/0
Chúc bạn học tốt
Bài 3 :
\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)
1 1 1 1
0,5 0,5 0,5 0,5
b) \(n_{H2}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)
c) \(n_{H2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{H2SO4}=0,5.98=49\left(g\right)\)
\(C_{ddH2SO4}=\dfrac{49.100}{200}=24,5\)0/0
d) \(n_{MgSO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{MgSO4}=0,5.120=60\left(g\right)\)
\(m_{ddspu}=12+200-\left(0,5.2\right)=211\left(g\right)\)
\(C_{MgSO4}=\dfrac{60.100}{211}=28,44\)0/0
Chúc bạn học tốt
\(a)n_{Al}=\dfrac{10,8}{27}=0,4mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,4 1,2 0,4 0,6
\(V_{H_2}=0,6.22,4=13,44l\\ b)m_{HCl}=1,2.36,5=43,8g\\ C_{\%HCl}=\dfrac{43,8}{200}\cdot100\%=21,9\%\)
hi Thắng tui là @Đinh Trí Gia Bình nek:)