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a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
______0,2---->0,3------------>0,1------>0,3______(mol)
=> VH2 = 0,3.22,4= 6,72(l)
b) \(C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,1}=3M\)
\(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,1}{0,1}=1M\)
Câu 3:
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=n_{H_2}=0,3(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,3.24}{15,2}.100\%=47,37\%\\ \Rightarrow \%_{MgO}=100\%-47,37\%=52,63\%\)
\(n_{MgO}=\dfrac{15,2-0,3.24}{40}=0,2(mol)\\ \Rightarrow \Sigma n_{HCl}=0,3.2+0,2.2=1(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{1.36,5}{10\%}=365(g)\\ \Sigma n_{MgCl_2}=0,2+0,3=0,5(mol)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{0,5.95}{15,2+365}.100\%=12,49\%\)
\(PTHH:Mg+2H_2SO_{4(đ)}\to MgSO_4+2H_2O+SO_2\uparrow\\ MgO+H_2SO_4\to MgSO_4+H_2O\\ \Rightarrow n_{SO_2}=n_{Mg}=0,3(mol)\\ \Rightarrow V_{SO_2}=0,3.22,4=6,72(l)\)
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)
$n_{Al_2(SO_4)_3} = 0,15(mol)$
$m_{dd\ sau\ pư} = 8,1 + 360 - 0,45.2 = 367,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{367,2}.100\% = 14\%$
Câu 3 :
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
1) Pt : \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,2 0,4 0,2
\(n_{MgCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{MgCl2}=0,2.136=27,2\left(g\right)\)
2) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,6.100}{20}=73\left(g\right)\)
Chúc bạn học tốt
Mình xin lỗi bạn nhé , bạn sửa lại giúp mình chỗ :
\(m_{MgCl2}=0,2.95=19\left(g\right)\)
Gọi $n_{Mg} = n_{Al} = a(mol) ; n_{Fe} = b(mol)$
Ta có :
$24a + 27a + 56b = 15,8(1)$
$n_{HCl} > 2n_{H_2}$ nên HCl dư
Ta có :
$n_{H_2} = a + 1,5a + b = \dfrac{13,44}{22,4} = 0,6(2)$
Từ (1)(2) suy ra a = 0,2 ; b = 0,1
$\%m_{Al} = \dfrac{0,2.27}{15,8}.100\% = 34,18\%$
$\%m_{Mg} = \dfrac{0,2.24}{15,8}.100\% = 30,38\%$
$\%m_{Fe} = 35,44\%$
$n_{HCl\ pư} = 2n_{H_2} = 1,2(mol)$
Bảo toàn khối lượng : $m_{muối} = 15,8 + 1,2.36,5 - 0,6.2 = 58,4(gam)$
a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)
c, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}\left(M\right)\)
\(C_{M_{FeCl_2}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)
c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)
d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)
e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)
d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)
Mg+ 2HCl→ MgCl2+ H2
(mol) 0,1 0,2 0,1 0,1
a) \(n_{Mg}=\dfrac{m}{M}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
→\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(lít\right)\)
b) Đổi: 100ml=0,1 lít
\(C_{M_{HCl}}=\dfrac{n}{V}=\dfrac{0,2}{0,1}=2M\)
c) \(m_{MgCl_2}=n.M=0,1.95=9,5\left(g\right)\)