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\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Pt: \(Mg+2HCl\rightarrow MgCl_2+H_2\) (1)
0,2mol \(\leftarrow\)0,4mol \(\leftarrow\)0,2mol \(\leftarrow\)0,2mol
\(m_{Mg}=0,2.24=4,8\left(g\right)\)
\(\Rightarrow m_{MgO}=8,8-4,8=4\left(g\right)\)
\(\%Mg=\dfrac{4,8}{8,8}.100=54,55\%\)
\(\%MgO=\dfrac{4}{8,8}.100=45,45\%\)
\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\) (2)
0,1mol \(\rightarrow\)0,2mol \(\rightarrow\)0,1mol
(1)(2) \(\Rightarrow\Sigma_{n_{HCl}}=0,4+0,2=0,6\left(mol\right)\)
\(m_{dd_{HCl}}=\dfrac{0,6.36,5}{7,3}.100=300\left(g\right)\)
\(\Sigma_{m_{dd\left(spu\right)\left(1\right)}}=4,8+300-0,2.2=304,4\left(g\right)\)
\(C\%_{MgCl_2\left(1\right)}=\dfrac{0,2.95}{304,4}.100=6,24\%\)
\(\Sigma_{m_{dd\left(spu\right)\left(2\right)}}=4+300=304\left(g\right)\)
\(C\%_{MgCl_2\left(2\right)}=\dfrac{0,1.95}{304}.100=3,125\%\)
\(a,n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Ba + 2H2O ---> Ba(OH)2 + H2
0,3<-------------0,3<---------0,3
=> mBa = 0,3.137 = 41,1 (g)
=> mK2O = 59,9 - 41,1 = 18,8 (g)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{41,1}{59,9}.100\%=68,61\%\\\%m_{K_2O}=100\%-68,61\%=31,39\%\end{matrix}\right.\)
\(b,n_{K_2O}=\dfrac{18,8}{94}=0,2\left(mol\right)\)
PTHH: K2O + H2O ---> 2KOH
0,2----------------->0,4
Các chất tan trong dd sau phản ứng: KOH, Ba(OH)2
\(\rightarrow\left\{{}\begin{matrix}m_{KOH}=0,4.56=22,4\left(g\right)\\m_{Ba\left(OH\right)_2}=0,3.171=51,3\left(g\right)\end{matrix}\right.\)
a,Fe + 2HCl → FeCl + H2 (1)
FeO + 2HCl → FeCl + H2O (2)
nH2 = 3,36/ 22,4 = 0,15 ( mol)
Theo (1) nH2 = nFe = 0,15 ( mol)
mFe = 0,15 x 56 = 8.4 (g)
m FeO = 12 - 8,4 = 3,6 (g)
a, \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl->FeCl_2+H_2\left(1\right)\)
\(FeO+2HCl->FeCl_2+H_2O\left(2\right)\)
theo (1) \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{FeO}=12-8,4=3,6\left(g\right)\)
ta thấy : nFe =nH2 = 0,15
=> mFe =0,15 x 56 = 8,4g
%Fe=8,4/12 x 100 = 70%
=>%FeO = 100 - 70 = 30%
b) BTKLra mdd tìm mct of HCl
c) tìm mdd sau pứ -mH2 nha bạn
\(Fe\left(0,1\right)+2HCl\left(0,2\right)--->FeCl_2\left(0,1\right)+H_2\left(0,1\right)\)\(\left(1\right)\)
\(Fe_2O_3\left(0,05\right)+6HCl\left(0,3\right)--->2FeCl_3\left(0,1\right)+3H_2O\)\(\left(2\right)\)
\(n_{H_2}=0,1\left(mol\right)\)
Theo (1) \(n_{Fe}=0,1\left(mol\right)\)\(\Rightarrow m_{Fe}=5,6\left(g\right)\)
\(\Rightarrow m_{Fe_2O_3}=13,6-5,6=8\left(g\right)\)\(\Rightarrow n_{Fe_2O_3}=0,05\left(mol\right)\)
=> Rainbow có: \(\left\{{}\begin{matrix}\%m_{Fe}=41,18\%\\\%m_{Fe_2O_3}=58,82\%\end{matrix}\right.\)
Phản ứng vừa đủ => Dung dịch thu được sau phản ứng : \(\left\{{}\begin{matrix}FeCl_2:0,1\left(mol\right)\\FeCl_3:0,1\left(mol\right)\end{matrix}\right.\)
Theo (1) và (2) \(\sum n_{HCl}=0,5\left(mol\right)\)
\(\Rightarrow m_{HCl}=18,25\left(g\right)\)\(\Rightarrow m_{ddHCl}=625\left(g\right)\)
\(m dd sau =13,6+625-0,1.2=638,4(g)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,1.127.100}{638,4}=1,99\%\\C\%_{FeCl_3}=\dfrac{0,1.162,5.100}{638,4}=2,55\%\end{matrix}\right.\)
Ta co pthh
Fe + 2HCl \(\rightarrow\) FeCl2 + H2 (1)
Fe2O3 + 6HCl \(\rightarrow\) 2FeCl3 + 3H2O (2)
Ta thấy trong 2 pt thì pt 1 sinh ra khí H2 nên
ta có
nH2=\(\dfrac{2,24}{22,4}=0,1mol\)
Theo pthh 1
nFe=nH2=0,1 mol
\(\rightarrow\) mFe=0,1.56=5,6 g
mFe2O3=13,6-5,6=8 g
a, Ta co
%mFe=\(\dfrac{5,6.100\%}{13,6}\approx41,18\%\)
%mFe2O3=100%-41,18%=58,2%
b, Theo pthh 1
nHCl=2nH2=2.0,1=0,2 mol
nFeCl2=nH2=0,1 mol -> mFeCl2=0,1.127=12,7 g
Theo pthh 2
nFe2O3=\(\dfrac{8}{160}=0,05mol\)
nFeCl3=2nFe2O3=2.0,05 = 0,1 mol -> mFeCl3=0,1.162,5 =16,25 g
nHCl=6nFe2O3=6.0,05=0,3 mol
-> số mol của HCl (1) và (2) sau phản ứng là :
nHCl(sau-phan-ung) = 0,2+0,3 = 0,5 mol
-> mct=mHCl=0,5 .36,5 = 18,25 g
mddHCl=\(\dfrac{mct.100\%}{C\%}=\dfrac{18,25.100\%}{2,92\%}=625\left(g\right)\)
-> mdd(sau-phan-ung) = m(hon-hop-ban-dau) + mddHCl(sau-phan-ung) - mH2 = 13,6 + 625 - (0,1.2) = 638,4 g
\(\Rightarrow\) C%\(_{\text{dd}FeCl2}=\dfrac{12,7}{638,4}.100\%\approx1,99\%\)
C%\(_{\text{dd}FeCl3}=\dfrac{16,25}{638,4}.100\%\approx2,55\%\)
\(n_{HCl}=\dfrac{500.7,3}{100}:36,5=1\left(mol\right)\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
x 2x x
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
y 6y 2y
Đặt \(n_{MgO}:x\left(mol\right),n_{Al_2O_3}:y\left(mol\right)\)
Có hệ \(\left\{{}\begin{matrix}2x+6y=1\\40x+102y=18,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow n_{MgCl_2}=x=0,2\left(mol\right);n_{AlCl_3}=2y=2.0,1=0,2\left(mol\right)\)
\(C\%_{MgCl_2}=\dfrac{0,2.95.100}{18,2+500}=3,67\%\)
\(C\%_{AlCl_3}=\dfrac{0,2.133,5.100}{18,2+500}=5,15\%\)
Bài 14 :
\(a) n_{CuO} = \dfrac{8}{80} = 0,1(mol)\\ CuO + 2HCl \to CuCl_2 + H_2O\\ n_{HCl} = 2n_{CuO} = 0,2(mol)\\ m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)\\ b) \text{Chất tan : } CuCl_2\\ n_{CuCl_2} = n_{CuO} = 0,1(mol)\\ m_{CuCl_2} = 0,1.135 = 13,5(gam)\)
Bài 15 :
\(a) n_{Fe_2O_3} =\dfrac{4,8}{160} = 0,03(mol)\\ Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O\\ n_{H_2SO_4} = 3n_{Fe_2O_3} = 0,09(mol)\\ m_{dd\ H_2SO_4} = \dfrac{0,09.98}{9,8\%} = 90(gam)\\ b) \text{Chất tan : } Fe_2(SO_4)_3\\ n_{Fe_2(SO_4)_3} = n_{Fe_2O_3} = 0,03(mol)\\ m_{Fe_2(SO_4)_3} = 0,03.400 = 12(gam)\)
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