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\(n_{H_2}=\dfrac{2,36}{22,4}=\dfrac{59}{560}\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ n_{Al}=\dfrac{2}{3}.\dfrac{59}{560}=\dfrac{59}{840}\left(mol\right)\\ \Rightarrow\%m_{Al}=\dfrac{\dfrac{59}{840}.27}{50}.100\approx3,793\%\\ \Rightarrow\%m_{Al_2O_3}\approx96,207\%\)
a) \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Mol: x 1,5x
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: y y
Ta có: \(\left\{{}\begin{matrix}27x+65y=24,9\\1,5x+y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}27x+65.\left(0,6-1,5x\right)=24,9\\y=0,6-1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)
b, \(m_{Al}=0,2.27=5,4\left(g\right);m_{Zn}=24,9-5,4=19,5\left(g\right)\)
c) \(\%m_{Al}=\dfrac{5,4.100\%}{24,9}=21,69\%;\%m_{Zn}=100\%-21,69\%=78,31\%\)
d)
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Mol: 0,2 0,6 0,2 0,3
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,3 0,6 0,3 0,3
\(m_{ddHCl}=\dfrac{\left(0,6+0,6\right).36,5.100}{14}=312,857\left(g\right)\)
e) mdd sau pứ = 24,9 + 312,857 - (0,3+0,3).2 = 336,557 (g)
\(C\%_{ddAlCl_3}=\dfrac{0,2.133,5.100\%}{336,557}=7,93\%\)
\(C\%_{ddZnCl_2}=\dfrac{0,3.136.100\%}{336,557}=12,12\%\)
\(a)n_{H_2}=\dfrac{7,437}{24,79}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,1 0,3
\(\%m_{Al}=\dfrac{0,2.27}{15,6}\cdot100=34,62\%\\ \%m_{Al_2O_3}=100-34,62=65,38\%\\ b)n_{Al_2O_3}=\dfrac{\left(15,6-0,2.27\right)}{102}=0,1mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
0,1 0,15 0,05 0,15
\(m_{ddH_2SO_4}=\dfrac{\left(0,3+0,15\right).98}{9,8}\cdot100=450g\)
c) \(C_{\%Al_2\left(SO_4\right)_3}=\dfrac{\left(0,1+0,05\right)342}{15,6+450-0,3.2}\cdot100=11\%\)
Mg+H2SO4->MgSO4+H2
2Al+3H2SO4->Al2(SO4)3+3H2
Fe+H2SO4->FeSO4+H2
Zn+H2SO4->ZnSO4+H2
m muối=m kim loại+m gốc axit
7.48=1.72+mSO4/H2SO4
mSO4=5.76(g)
nSO4/H2SO4=0.06(mol)
->nH2SO4=0.06(mol)
Theo pthh 1 2 3 4
nH2SO4:nH2=1
->nH2SO4=nH2->nH2=0.06(mol)
V=0.06*22.4=1.344
a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)
1 2 1 1
a 2a
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
1 6 2 3
b 6b
b) Gọi a là số mol của CuO
b là số mol của Fe2O3
\(m_{CuO}+m_{Fe2O3}=25\left(g\right)\)
⇒ \(n_{CuO}.M_{CuO}+n_{Fe2O3}.M_{Fe2O3}=25g\)
⇒ 80a + 160b = 25g (1)
Ta có : 200ml = 0,2l
\(n_{HCl}=3,5.0,2=0,7\left(mol\right)\)
⇒ 2a + 6b = 0,7(2)
Từ (1),(2) ta có hệ phương trình :
80a + 160b = 25g
2a + 6b = 0,7
⇒ \(\left\{{}\begin{matrix}a=0,2375\\b=0,0375\end{matrix}\right.\)
\(m_{CuO}=0,2375.80=19\left(g\right)\)
\(m_{Fe2O3}=0,0375.160=6\left(g\right)\)
Chúc bạn học tốt
\(n_{H_2}=\dfrac{0.56}{22.4}=0.025\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(n_{Al}=\dfrac{2}{3}\cdot0.025=\dfrac{1}{60}\left(mol\right)\)
\(m_{Al}=\dfrac{1}{60}\cdot27=0.45\left(g\right)\)
\(m_{Cu}=25-0.45=24.55\left(g\right)\)
\(\%Cu=\dfrac{24.55}{25}\cdot100\%=98.2\%\)
\(\%Al=100-98.2=1.8\%\)
\(Cu+2H_2SO_{4\left(đ\right)}\rightarrow CuSO_4+SO_2+2H_2O\)
\(2Al+6H_2SO_{4\left(đ\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)
\(n_{Cu}=\dfrac{24.55}{64}=\dfrac{491}{1280}\left(mol\right)\)
\(V_{SO_2}=\left(\dfrac{1}{60}\cdot\dfrac{3}{2}+\dfrac{491}{1280}\right)\cdot22.4=9.1525\left(l\right)\)
a)
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
b)
$n_{H_2} = \dfrac{0,56}{22,4} = 0,225(mol)$
Theo PTHH : $n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{1}{60}(mol)$
$m_{Al} = \dfrac{1}{60}.27 = 0,45(gam)$
$m_{Cu} = 25 - 0,45 = 24,55(gam)$
c)
$\%m_{Al} = \dfrac{0,45}{25}.100\% = 1,8\%$
$\%m_{Cu} = 100\% -1,8\% = 98,2\%$
d)
$Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$
$2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O$
Theo PTHH :
$n_{SO_2} = n_{Cu} + \dfrac{3}{2}n_{Al} = \dfrac{24,55}{64} + \dfrac{1}{60}.\dfrac{3}{2} = 0,41(mol)$
$V_{SO_2} = 0,41.22,4 = 9,184(lít)$