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a) 2Al + 6HCl -> 2AlCl3 + 3H2
Al2O3 + 6HCl -> 2AlCl3 + 3H2O
nH2 = 0,15mol => nAl=0,1mol => mAl=2,7g; mAl2O3 = 10,2g => nAl2O3 = 0,1mol
=>%mAl=20,93% =>%mAl2O3 = 79,07%
b) nHCl = 0,1.3+0,1.6=0,9 mol=>mHCl(dd)=100g
mddY=12,9+100-0,15.2=112,6g
mAlCl3=22,5g=>C%=19,98%
Câu 1:
Gọi số mol NaCl, KCl là a, b (mol)
=> 58,5a + 74,5b = 6,81 (1)
\(n_{AgCl}=\dfrac{14,35}{143,5}=0,1\left(mol\right)\)
Bảo toàn Cl: a + b = 0,1 (2)
(1)(2) => a = 0,04 (mol); b = 0,06 (mol)
\(\left\{{}\begin{matrix}m_{NaCl}=0,04.58,5=2,34\left(g\right)\\m_{KCl}=0,06.74,5=4,47\left(g\right)\end{matrix}\right.\)
Câu 2:
Gọi số mol MgCl2, KCl là a, b (mol)
=> 95a + 74,5b = 3,93 (1)
25ml dd A chứa \(\left\{{}\begin{matrix}MgCl_2:0,05a\left(mol\right)\\KCl:0,05b\left(mol\right)\end{matrix}\right.\)
nAgNO3 = 0,05.0,06 = 0,003 (mol)
=> nAgCl = 0,003 (mol)
Bảo toàn Cl: 0,1a + 0,05b = 0,003 (2)
(1)(2) => a = 0,01 (mol); b = 0,04 (mol)
\(\left\{{}\begin{matrix}\%m_{MgCl_2}=\dfrac{0,01.95}{3,93}.100\%=24,173\%\\\%m_{KCl}=\dfrac{0,04.74,5}{3,93}.100\%=75,827\%\end{matrix}\right.\)
a) Gọi số mol Mg, CuO là a, b (mol)
=> 24a + 80b = 14 (1)
\(n_{HCl}=\dfrac{255,5.10\%}{36,5}=0,7\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
a--->2a--------->a------>a
CuO + 2HCl --> CuCl2 + H2O
b------>2b----->b
=> 2a + 2b = 0,7 (2)
(1)(2) => a = 0,25 (mol); b = 0,1 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25.24}{14}.100\%=42,857\%\\\%m_{CuO}=\dfrac{0,1.80}{14}.100\%=57,143\%\end{matrix}\right.\)
b)
mdd sau pư = 14 + 255,5 - 0,25.2 = 269 (g)
\(C\%_{MgCl_2}=\dfrac{0,25.95}{269}.100\%=8,829\%\)
\(C\%_{CuCl_2}=\dfrac{0,1.135}{269}.100\%=5,019\%\)
Chất rắ B là Cu
Fe3O4+8HCl--->2FeCl2+FeCl3+4H2O
FeCl2+2AgNO3--->Fe(NO3)2+2AgCl
FeCl3+3AgNO3--->Fe(NO3))3+3AgCl
n FE3O4=23,2/232=0,1(mol)
n HCl=0,4.2=0,8(mol)
Do 0,1/1=0,8/1--> sau pư ko có dd HCl dư
Theo pthh1
n FECl3=2n Fe3O4=0,2(mol)
Theo pthh3
n AgCl=3n FeCl3=0,6(mol)
Theo pthh1
n FeCl2=n Fe3O4=0,1(mol)
Theo pthh2
n AgCl=2n FeCl2=0,2(mol)
Tổng n AgCl=0,2+0,6=0,8(mol)
m X=m AgCl=143,5.0,8=114,8(g)
\(Đặt:n_{MnO_2}=a\left(mol\right),n_{KMnO_4}=b\left(mol\right)\)
\(m_{hh}=87a+158b=37.96\left(g\right)\left(1\right)\)
\(n_{Cl_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
\(n_{Cl_2}=a+2.5b=0.45\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.4,b=0.02\)
\(\%MnO_2=\dfrac{0.4\cdot87}{37.96}\cdot100\%=91.68\%\\\%KMnO_4=100-91.68=8.32\% \)
\(m_M=m_{KCl}+m_{MnCl_2}=0.02\cdot74.5+\left(0.4+0.02\right)\cdot126=54.41g\)
\(a,n_{Cl_2}=\dfrac{15,62}{71}=0,22\left(mol\right)\\ Đặt:n_{Fe}=a\left(mol\right);n_{Cu}=b\left(mol\right)\left(a,b>0\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ Cu+Cl_2\rightarrow\left(t^o\right)CuCl_2\\ \Rightarrow\left\{{}\begin{matrix}56a+64b=10,88\\1,5a+b=0,22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,08\\b=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Fe}=\dfrac{0,08.56}{10,88}.100\approx41,176\%\Rightarrow\%m_{Cu}\approx58,824\%\\ b,FeCl_3+3AgNO_3\rightarrow Fe\left(NO_3\right)_3+3AgCl\downarrow\\ CuCl_2+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2AgCl\\ n_{AgNO_3}=3.0,08+2.0,1=0,44\left(mol\right)\\ m_{AgNO_3}=0,24.170=40,8\left(g\right)\\ m_{ddAgNO_3}=\dfrac{40,8}{24\%}=170\left(g\right)\)
\(c,2NaI+Cl_2\rightarrow2NaCl+I_2\\ n_{NaI}=2.n_{Cl_2}=2.0,22=0,44\left(mol\right)\Rightarrow a=C_{MddNaI}=\dfrac{0,44}{0,1}=4,4\left(M\right)\\ n_{I_2}=n_{Cl_2}=0,22\left(mol\right)\Rightarrow m_{I_2}=254.0,22=55,88\left(g\right)\)
\(d,MnO_2+4HCl_{\left(đặc\right)}\rightarrow\left(t^o\right)MnCl_2+Cl_2+2H_2O\\ n_{MnO_2}=n_{Cl_2}=0,22\left(mol\right)\\ \Rightarrow x=m_{MnO_2}=87.0,22=19,14\left(g\right)\\ n_{HCl}=4.0,22=0,88\left(mol\right)\\ y=C_{MddHCl}=\dfrac{0,88}{0,05}=17,6\left(M\right)\)
a)
Gọi : \(\left\{{}\begin{matrix}n_{NaCl}=a\left(mol\right)\\n_{KI}=b\left(mol\right)\end{matrix}\right.\)
NaCl + AgNO3 → AgCl + NaNO3
a..............a...............a..............................(mol)
KI + AgNO3→ AgI + KNO3
b.......b..............b..................................(mol)
Ta có :
\(n_{AgNO_3} = a + b = 0,25.2 = 0,5(mol)\)
\(m_{kết\ tủa} = 143,5a + 235b = 103,775\)(gam)
Suy ra : a = 0,15 ; b = 0,35
Vậy :
\(C_{M_{NaCl}} = \dfrac{0,15}{0,4} = 0,375M\\ C_{M_{KI}} = \dfrac{0,35}{0,4} = 0,875M\)
b)
Sau phản ứng, dung dịch gồm : \(\left\{{}\begin{matrix}NaNO_3:0,15\left(mol\right)\\KNO_3:0,35\left(mol\right)\end{matrix}\right.\)
Suy ra :
\(m_{NaNO_3} = 0,15.85 = 12,75(gam)\\ m_{KNO_3} = 0,35.101 = 35,35(gam)\)