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a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
_____0,2-->0,6----->0,2---->0,3
=> \(C_{M\left(HCl\right)}=\dfrac{0,6}{0,2}=3M\)
b) VH2 = 0,3.24,79 = 7,437 (l)
c) \(C_{M\left(AlCl_3\right)}=\dfrac{0,2}{0,2}=1M\)
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.2.......0.4........0.2.......0.2\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(V_{dd_{HCl}}=\dfrac{0.4}{0.2}=2\left(l\right)\)
\(C_{M_{FeCl_2}}=\dfrac{0.2}{2}=0.1\left(M\right)\)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{H_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$V_{H_2} = 0,2.22,4 = 4,48(lít)$
b)
$n_{HCl} = 2n_{Fe} = 0,4(mol)$
$V_{dd\ HCl} = \dfrac{0,4}{0,2} = 2(lít) = 2000(ml)$
c)
$n_{FeCl_2} = n_{Fe} = 0,2(mol)$
$\Rightarrow C_{M_{FeCl_2}} = \dfrac{0,2}{2} = 0,1M$
2Al+ 3H2SO4→ Al2(SO4)3+ 3H2
(mol) 0,1 0,15 0,05 0,15
đổi: 300ml=0,3 lít
a) nAl=\(\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=C_M.V=0,5.0,3=0,15\left(mol\right)\)
tỉ lệ:
Al H2SO4
\(\dfrac{0,2}{2}\) > \(\dfrac{0,15}{3}\)
→ Al dư, H2SO4 phản ứng hết sau phản ứng
→ \(V_{H_2}=n.22,4=0,15.22,4=3,36\left(lít\right)\)
b) \(n_{Al\left(dư\right)}=n_{Al\left(bđ\right)}-n_{Al\left(ph.ứ\right)}=0,2-0,1=0,1\left(mol\right)\)
\(C_{M_{Al\left(dư\right)}}=\dfrac{n}{V}=\dfrac{0,1}{0,3}=\dfrac{1}{3}M\)
\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{n}{V}=\dfrac{0,05}{0,3}=\dfrac{1}{6}M\)
\(C_{M_{H_2}}=\dfrac{n}{V}=\dfrac{0,15}{0,3}=0,05M\)
a, \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
b, \(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
Theo PT: \(n_{CuO}=n_{CuCl_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,05.80=4\left(g\right)\)
c, \(C_{M_{CuCl_2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
PTHH :
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,05 0,1 0,05
\(b,m_{CuO}=0,05.80=4\left(g\right)\)
\(c,C_{M\left(CuCl_2\right)}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(n_{Al}=\frac{2,7}{27}=0,1\left(mol\right)\)
\(n_{Al_2O_3}=\frac{5,1}{102}=0,05\left(mol\right)\)
2Al + 3H2SO4 → Al2(SO4)3 + 3H2 (1)
Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O (2)
a) Theo PT1: \(n_{H_2}=\frac{3}{2}n_{Al}=\frac{3}{2}\times0,1=0,15\left(mol\right)\)
\(\Rightarrow V=V_{H_2}=0,15\times22,4=3,36\left(l\right)\)
b) Theo Pt1: \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)
Theo pT2: \(n_{H_2SO_4}=3n_{Al_2O_3}=3\times0,05=0,15\left(mol\right)\)
\(\Rightarrow\Sigma n_{H_2SO_4}=0,15+0,15=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,3\times98=29,4\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{29,4}{9,8\%}=300\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\frac{300}{1,12}=267,86\left(ml\right)=0,26786\left(l\right)\)
c) Ta có: \(m_{ddY}=2,7+5,1+300=307,8\left(g\right)\)
Theo PT1: \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{2}n_{Al}=\frac{1}{2}\times0,1=0,05\left(mol\right)\)
Theo Pt2: \(n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,05\left(mol\right)\)
\(\Rightarrow\Sigma n_{Al_2\left(SO_4\right)_3}=0,05+0,05=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1\times342=34,2\left(g\right)\)
\(\Rightarrow C_{M_{Al_2\left(SO_4\right)_3}}=\frac{0,1}{0,26786}=0,373\left(M\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\frac{34,2}{307,8}\times100\%=11,11\%\)