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Cl2+2NaBr ->2NaCl+Br2
0,04 -0,08-------0,08-----0,04 mol
Cl2+2NaI ->2NaCl+I2
0,03--0,06---0,06-----0,03 mol
n NaBr =0,8.0,1=0,08 mol
n NaI=0,6.0,1=0,06 mol
=>V Cl2=(0,04+0,03).22,4=1,568l
=>m Br2=0,04 .160=6,4g
=>m I2=0,03.254=7,62g
b)
NaCl+AgNO3->AgCl+NaNO3
0,14-----------------0,14 mol
=>m AgCl=0,14.143,5=20,09g
c)2KMnO4+16HCl->2KCl+2MnCl2+5Cl2+8H2O
0,028--------------------------------------0,07 mol
=>m KMnO4=0,028.158=4,424g
nAl = 5,4 / 27 = 0,2(mol)
2Al + 3H2SO4--- > Al2(SO4)3 + 3H2
0,2 0,3 0,1 0,3 (mol)
VH2SO4 = n/ CM = 0,3 / 2 = 0,15(l)
=> V1 = 150 ml
mAl2(SO4)3 = 0,1 . 342 = 34,2 (g)
Al2(SO4)3 + 3BaCl2 -- > 3BaSO4 + 2AlCl3
0,1 0,1
=> mBaSO4 = 0,1 . 233 = 23,3 (g)
Ta thấy khi cho Br 2 vào dung dịch 2 muối S 4 + thì toàn bộ S 4 + sẽ bị oxi hoá lên S 6 + ( SO 4 2 - )do đó :
n SO 2 = n SO 4 2 - = 0,15
=> m BaSO 4 = 0,15.233 = 34,95g
a)
Gọi số mol Mg, Al là a, b (mol)
=> 24a + 27b = 26,25 (1)
\(n_{H_2}=\dfrac{30,8}{22,4}=1,375\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
a-->2a--------->a------>a
2Al + 6HCl --> 2AlCl3 + 3H2
b---->3b------->b------>1,5b
=> a + 1,5b = 1,375 (2)
(1)(2) => a = 0,25 (mol); b = 0,75 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25.24}{26,25}.100\%=22,857\%\\\%m_{Al}=\dfrac{0,75.27}{26,25}.100\%=77,143\%\end{matrix}\right.\)
b)
nHCl = 2a + 3b = 2,75 (mol)
=> mHCl = 2,75.36,5 = 100,375 (g)
=> \(m_{dd.HCl}=\dfrac{100,375.100}{10}=1003,75\left(g\right)\)
c)
mdd sau pư = 1003,75 + 26,25 - 1,375.2 = 1027,25 (g)
\(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,25.95}{1027,25}.100\%=2,312\%\\C\%_{AlCl_3}=\dfrac{0,75.133,5}{1027,25}.100\%=9,747\%\end{matrix}\right.\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\n_{HCl}=2n_{Fe}=0,4\left(mol\right)\end{matrix}\right.\)
a, Ta có: \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(V_{ddHCl}=\dfrac{0,4}{1,5}\approx0,267\left(l\right)\)
c, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)
Bạn tham khảo nhé!
\(a,n_{Cl_2}=\dfrac{15,62}{71}=0,22\left(mol\right)\\ Đặt:n_{Fe}=a\left(mol\right);n_{Cu}=b\left(mol\right)\left(a,b>0\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ Cu+Cl_2\rightarrow\left(t^o\right)CuCl_2\\ \Rightarrow\left\{{}\begin{matrix}56a+64b=10,88\\1,5a+b=0,22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,08\\b=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Fe}=\dfrac{0,08.56}{10,88}.100\approx41,176\%\Rightarrow\%m_{Cu}\approx58,824\%\\ b,FeCl_3+3AgNO_3\rightarrow Fe\left(NO_3\right)_3+3AgCl\downarrow\\ CuCl_2+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2AgCl\\ n_{AgNO_3}=3.0,08+2.0,1=0,44\left(mol\right)\\ m_{AgNO_3}=0,24.170=40,8\left(g\right)\\ m_{ddAgNO_3}=\dfrac{40,8}{24\%}=170\left(g\right)\)
\(c,2NaI+Cl_2\rightarrow2NaCl+I_2\\ n_{NaI}=2.n_{Cl_2}=2.0,22=0,44\left(mol\right)\Rightarrow a=C_{MddNaI}=\dfrac{0,44}{0,1}=4,4\left(M\right)\\ n_{I_2}=n_{Cl_2}=0,22\left(mol\right)\Rightarrow m_{I_2}=254.0,22=55,88\left(g\right)\)
\(d,MnO_2+4HCl_{\left(đặc\right)}\rightarrow\left(t^o\right)MnCl_2+Cl_2+2H_2O\\ n_{MnO_2}=n_{Cl_2}=0,22\left(mol\right)\\ \Rightarrow x=m_{MnO_2}=87.0,22=19,14\left(g\right)\\ n_{HCl}=4.0,22=0,88\left(mol\right)\\ y=C_{MddHCl}=\dfrac{0,88}{0,05}=17,6\left(M\right)\)
\(a,PTHH:2Al+3X_2\rightarrow2AlX_3\\ Theo.ĐLBTKL,ta.có:\\ m_{Al}+m_{X_2}=m_{AlX_3}\\ \Leftrightarrow m_{Al}+33,6=37,38\\ \Leftrightarrow m_{Al}=3,78\left(g\right)\\ \Rightarrow n_{Al}=\dfrac{3,78}{27}=0,14\left(mol\right)\\ n_{X_2}=\dfrac{3}{2}.0,14=0,21\left(mol\right)\\ \Rightarrow M_{X_2}=\dfrac{33,6}{0,21}=160\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_X=80\left(\dfrac{g}{mol}\right)\\ \Rightarrow X:Brom\left(Br=80\right)\\ b,n_{AlBr_3}=\dfrac{42,72}{267}=0,16\left(mol\right)\\ AlBr_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgBr\downarrow\left(vàng.nhạt\right)\\ n_{AgBr}=n_{AgNO_3}=0,16.3=0,48\left(mol\right)\\ \Rightarrow m_T=m_{\downarrow}=m_{AgBr}=188.0,48=90,24\left(g\right)\)
\(m_{ddZ}=42,72+447,52-90,24=400\left(g\right)\\ n_{Al\left(NO_3\right)_3}=n_{AlBr_3}=0,16\left(mol\right)\\ \Rightarrow C\%_{ddZ}=C\%_{ddAl\left(NO_3\right)_3}=\dfrac{0,16.213}{400}.100=8,52\%\)
\(c,n_{Br_2}=\dfrac{12,8}{160}=0,08\left(mol\right)\\ 2NaI+Br_2\rightarrow2NaBr+I_2\\ n_{I_2}=n_{Br_2}=0,08\left(mol\right);n_{NaI}=2.0,08=0,16\left(mol\right)\\ \Rightarrow x=C_{MddNaI}=\dfrac{0,16}{0,25}=0,64\left(M\right)\\ m_{I_2}=0,08.254=20,32\left(g\right)\)
\(a,n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right);n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ Mg+Cl_2\rightarrow\left(t^o\right)MgCl_2\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ n_{Cl_2}=n_{Mg}+1,5.n_{Al}=0,1+1,5.0,1=0,25\left(mol\right)\\ \Rightarrow V=V_{Cl_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\\ b,MgCl_2+2KOH\rightarrow Mg\left(OH\right)_2\downarrow+2KCl\\ AlCl_3+3KOH\rightarrow Al\left(OH\right)_3\downarrow+3KCl\\KT.max\Leftrightarrow Al\left(OH\right)_3.không.tan.trong.kiềm\\ n_{Mg\left(OH\right)_2}=n_{MgCl_2}=n_{Mg}=0,1\left(mol\right);n_{Al\left(OH\right)_3}=n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\ \Rightarrow m_{\downarrow\left(max\right)}=0,1.58+0,1.78=13,6\left(g\right)\)
\(n_{KOH}=2.0,1+3.0,1=0,5\left(mol\right)\\ \Rightarrow m_{KOH}=0,5.56=28\left(g\right)\\ m_{ddKOH}=\dfrac{28.100}{4}=700\left(g\right)\\ c,MnO_2+4HCl_{đặc,nóng}\rightarrow MnCl_2+Cl_2+2H_2O\\ n_{MnO_2}=n_{Cl_2}=0,25\left(mol\right)\\ \Rightarrow m_{MnO_2}=0,25.87=21,75\left(g\right)\)