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\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.2.........0.4.........0.2......0.2\)
\(m_{Zn}=0.2\cdot65=13\left(g\right)\Rightarrow m_{ZnO}=14.6-13=1.6\left(g\right)\)
\(\%Zn=\dfrac{13}{14.6}\cdot100\%=89.04\%\)
\(\%ZnO=100\%-89.04\%=10.96\%\)
\(n_{ZnO}=\dfrac{1.6}{81}\approx0.02\left(mol\right)\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
\(0.02........0.04........0.02........0.02\)
\(n_{HCl}=0.4+0.04=0.44\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{0.44}{0.8}=0.55\left(M\right)\)
Eeeee ngồi tính sang chấn thật nó ra số xấu lần mò hơn 20p chưa biết tính sai chỗ nào
CuO + 2HCl -> CuCl2 + H2O (1)
Fe2O3 + 6HCl -> 2FeCl3 + 3H2O (2)
nHCl=0,2.3,5=0,7(mol)
Đặt nCuO=a
nFe2O3=b
Ta có hệ:
80a+160b=20
2a+6b=0,7
=>a=0,05;b=0,1
mCuO=80.0,05=4(g)
mFe2O3=20-4=16(g)
Theo PTHH 1 và 2 ta có:
nCuCl2=nCuO=0,05(mol)
nFeCl3=2nFe2O3=0,2(mol)
mCuCl2=135.0,05=6,75(g)
mFeCl3=162,5.0,2=32,5(g)
mdd =20+200.1,1=240(g)
C% dd CuCl2=6,72\240 .100%=2,8125%
C% dd FeCl3= 32,5\240 .100%=13,54%
Sửa đề: Sau phản ứng thu đc \(2240(cm^3)\) lít khí (đktc)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ ZnO+2HCl\to ZnCl_2+H_2O\\ b,n_{Zn}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Zn}=0,1.65=6,5(g)\\ \Rightarrow \%_{Zn}=\dfrac{6,5}{14,6}.100\%= 44,52\%\\ \Rightarrow \%_{ZnO}=100\%-44,52\%=55,48\%\\ n_{ZnO}=\dfrac{14,6-6,5}{81}=0,1(mol)\\ \Sigma n_{ZnCl_2}=n_{Zn}+n_{ZnO}=0,1+0,1=0,2(mol)\\ \Rightarrow C_{M_{ZnCl_2}}=\dfrac{0,2}{0,2}=1M\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a, Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)
⇒ 24x + 27y = 12,6 (1)
Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y\left(mol\right)\)
\(\Rightarrow x+\dfrac{3}{2}y=0,6\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,3\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MG}=\dfrac{0,3.24}{12,6}.100\%\approx57,1\%\\\%m_{Al}\approx42,9\%\end{matrix}\right.\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4}=n_{H_2}=0,6\left(mol\right)\\n_{MgSO_4}=n_{Mg}=0,3\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{H_2SO_4}=0,6.98=58,8\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{58,8}{14,7\%}=400\left(g\right)\)
Ta có: m dd sau pư = 12,6 + 400 - 0,6.2 = 411,4 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{0,3.120}{411,4}.100\%\approx8,75\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{411,4}.100\%\approx8,31\%\end{matrix}\right.\)
Bạn tham khảo nhé!
PTHH: \(CaO+2HCl\rightarrow CaCl_2+H_2O\) (1)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\) (2)
a) Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)=n_{CaCO_3}\)
\(\Rightarrow m_{CaCO_3}=0,2\cdot100=20\left(g\right)\) \(\Rightarrow\%m_{CaCO_3}=\dfrac{20}{25,6}\cdot100\%=78,125\%\)
\(\Rightarrow\%m_{CaO}=21,875\%\)
b) Theo 2 PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(2\right)}=2n_{CaCO_3}=0,4mol\\n_{HCl\left(1\right)}=2n_{CaO}=2\cdot\dfrac{25,6-20}{56}=0,2mol\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{HCl}=0,6mol\) \(\Rightarrow C\%_{HCl}=\dfrac{0,6\cdot36,5}{210\cdot1,05}\cdot100\%\approx9,93\%\)
\(n_{FeO}=a\left(mol\right),n_{CuO}=b\left(mol\right)\)
\(m_{hh}=72a+80b=19.2\left(g\right)\left(1\right)\)
\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(n_{H_2SO_4}=a+b=0.25\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.15\)
\(m_{FeO}=0.1\cdot72=7.2\left(g\right)\)
\(m_{CuO}=12\left(g\right)\)
\(C_{M_{FeSO_4}}=\dfrac{0.1}{0.25}=0.4\left(M\right)\)
\(C_{M_{CuSO_4}}=\dfrac{0.15}{0.25}=0.6\left(M\right)\)
a, PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
Ta có: 100nCaCO3 + 84nMgCO3 = 14,2 (1)
Theo PT: \(n_{CO_2}=n_{CaCO_3}+n_{MgCO_3}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CaCO_3}=0,1\left(mol\right)\\n_{MgCO_3}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CaCO_3}=\dfrac{0,1.100}{14,2}.100\%\approx70,42\%\\\%m_{MgCO_3}\approx29,58\%\end{matrix}\right.\)
b, Theo PT: \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,6}=0,5\left(M\right)\)
PTHH :
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
x 2x x x x
\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\uparrow\)
y 2y y y y
Có:
\(\left\{{}\begin{matrix}100x+84y=14,2\\x+y=\dfrac{3,36}{22,4}=0,15\end{matrix}\right.\)
\(\Rightarrow x=0,1;y=0,05\)
\(a,\%m_{CaCO_3}=0,1.100:14,2.100\%\approx72,423\%\)
\(\%m_{MgCO_3}=100\%-72,423\%\approx29,577\%\)
\(b,C_{M\left(HCl\right)}=\dfrac{0,2+0,1}{0,6}=0,5\left(M\right)\)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH :
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,25 0,5 0,25
\(a,m_{Fe}=0,25.56=14\left(g\right)\)
\(b,C_{M\left(HCl\right)}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)
\(a)n_{HCl}=0,2.1,5=0,3mol\\ CaO+2HCl\rightarrow CaCl_2+H_2O\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}2n_{CaO}+2n_{CuO}=0,3\\56n_{CaO}+80n_{CuO}=10,8\end{matrix}\right.\\ \Rightarrow n_{CaO}=n_{CaCl_2}=0,05mol;n_{CuO}=n_{CuCl_2}=0,1mol\\ \%m_{CaO}=\dfrac{0,05.56}{10,8}\cdot100=25,93\%\\ \%m_{CuO}=100-25,93=74,07\%\\ b)C_{M_{CaCl_2}}=\dfrac{0,05}{0,2}=0,25M\\ C_{M_{CuCl_2}}=\dfrac{0,1}{0,2}=0,5M\)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
x 2x x x
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
y 2y y y
\(\left\{{}\begin{matrix}56x+80y=10,8\\2x+2y=0,2.1,5=0,3\end{matrix}\right.\)
\(\Rightarrow x=0,05;y=0,1\)
\(a,\%m_{CaO}=0,05.56:10,8.100\%=25,93\left(\%\right)\)
\(\%m_{CuO}=100\%-25,93\%=74,07\%\)
\(b,C_{M\left(CaCl_2\right)}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(C_{M\left(CuCl_2\right)}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)