Hóa học

 \(Gọi.C_{M\left(H_2SO_4\right)}=a\left(M\right);C_{M\left(NaOH\right)}=b\left(M\right)\\ +TN1:\\ n_{H_2SO_4}=0,03a\left(mol\right);n_{NaOH}=0,02b\left(mol\right);n_{KOH}=0,01\cdot2=0,02mol\\ n_{H^{^+}}=0,03a\cdot2=0,06b\left(mol\right)\\ n_{\left(OH\right)^{^-}}=0,02b+0,02\left(mol\right)\\ H^{^+}+OH^{^-}->H_2O\\ \Rightarrow0,06a=0,02b+0,02\left(I\right)\\ +TN2:\\ n_{NaOH}=0,03b\left(mol\right);n_{H_2SO_4}=0,02a\left(mol\right)\\ n_{HCl}=5\cdot10^{^{-3}}mol\\ \Rightarrow n_{OH^{^-}}=0,03b\left(mol\right);n_{H^{^+}}=0,04a+5\cdot10^{^{-3}}\left(mol\right)\\ H^{^+}+OH^{^-}->H_2O\)

\(\Rightarrow0,03b=0,04a+5\cdot10^{^{-3}}\left(II\right)\\ \left(I\right)\left(II\right)\Rightarrow a=0,7M;b=1,1M\)

\(a.n_{CuO}=\dfrac{16}{80}=0,2mol\\ CuO+H_2SO_4->CuSO_4+H_2O\\ m_{ddH_2SO_4}=\dfrac{0,2\cdot98\cdot100}{10}=196\left(g\right)\\ C_{\%\left(CuSO_4\right)}=\dfrac{0,2\cdot160}{196+16}\cdot100\%=15,09\%\\ b.CuSO_4+2NaOH->Cu\left(OH\right)_2+Na_2SO_4\\ n_{NaOH}=2\cdot0,2=0,4mol\\ V_{ddNaOH}=\dfrac{0.4}{2}=0,2\left(L\right)=200\left(mL\right)\\ m_{kt}=98\cdot0,2=19,6g\)

\(n_{CuO}=\dfrac{8}{80}=0,1mol\\ n_{H_2SO_4}=\dfrac{20\%\cdot125}{98}=0,255mol\\ a.CuO+H_2SO_4->CuSO_4+H_2O\\ m_{H_2SO_4pư}=0,1\cdot98=9,8g\\ b.Chất.sau.pư.gồm.H_2SO_4dư,CuSO_4\\ C_{\%\left(H_2SO_4\right)}=\dfrac{98\left(0,255-0,1\right)}{133}\cdot100\%=11,42\%\\ C_{\%\left(CuSO_4\right)}=\dfrac{160\cdot0,1}{133}\cdot100\%=12,03\%\)

`n_[Fe_2 O_3]=24/160=0,15(mol)`

`Fe_2 O_3 + 3H_2 SO_4 -> Fe_2(SO_4)_3 + 3H_2 O`

   `0,15`              `0,45`                    `0,15`                                `(mol)`

`a)m_[Fe_2(SO_4)_3]=0,15.400=60(g)`

`b)C%_[Fe_2(SO_4)_3]=60/[24+100].100=48,39%`

`c)C%_[H_2 SO_4]=[0,45.98]/100 .100=44,1%`

\(a.Fe_2O_3+3H_2SO_4->Fe_2\left(SO_4\right)_3+3H_2O\\ m_{muối}=400\cdot\dfrac{24}{160}=60\left(g\right)\\ b.C_{\%}=\dfrac{60}{124}\cdot100\%=48,39\%\\ c.C_{\%}=\dfrac{3\cdot\dfrac{24}{60}\cdot98}{124}\cdot100\%=94,84\%\)

\(a,Na_2O+H_2O\rightarrow2NaOH\\ a,n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\\ n_{NaOH}=2.0,25=0,5\left(mol\right)\\ C_{MddA}=C_{MddNaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ b,2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{ddH_2SO_4}=\dfrac{0,25.98.100}{20}=122,5\left(g\right)\\ V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,456\left(ml\right)\)

Em đăng bài vào môn Toán nhé!

\(Cu\left(NO_3\right)_2\rightarrow\left(t^o\right)CuO+2NO_2+\dfrac{1}{2}O_2\\ ĐLBTKL:m_{Cu\left(NO_3\right)_2\left(p.ứ\right)}=m_{CuO}+m_{NO_2}+m_{O_2}=160+184+32=376\left(g\right)\\ b,\%m_{\dfrac{Cu\left(NO_3\right)_2\left(p.ứ\right)}{Cu\left(NO_3\right)_2\left(bđ\right)}}=\dfrac{376}{500}.100\%=75,2\%\)

\(a,2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ b,n_{H_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\Rightarrow V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{HCl}=\dfrac{6}{2}.0,2=0,6\left(mol\right)\\ m_{ddHCl}=\dfrac{0,6.36,5.100}{8,76}=250\left(g\right)\\ d,n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\Rightarrow m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\\ m_{ddsau}=5,4+250-0,3.2=254,8\left(g\right)\\ C\%_{ddAlCl_3}=\dfrac{26,7}{254,8}.100\approx10,479\%\)

\(n_{Al}=\dfrac{5,4}{27}=0,2mol\\ a.2Al+6HCl->2AlCl_3+3H_2\\ b.n_{H_2}=\dfrac{3}{2}\cdot0,2=0,3mol\\ V_{H_2}=0,3\cdot22,4=6,72\left(L\right)\\ c.n_{HCl}=3n_{Al}=0,6mol\\ V_{ddHCl}=\dfrac{0,6\cdot36,5\cdot100}{8,76}=250\left(g\right)\\ d.n_{AlCl_3}=n_{Al}=0,2mol\\ m_{muối}=0,2\cdot133,5=26,7\left(g\right)\\ e.m_{ddspu}=5,4+250-0,3\cdot2=254,8\left(g\right)\\ C_{\%\left(AlCl_3\right)}=\dfrac{26,7}{254,8}\cdot100\%=10,48\%\)

a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

b) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{H_2}=\dfrac{0,2.3}{2}=0,3\left(mol\right)\)

\(V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)

c) \(n_{HCl}=\dfrac{0,2.6}{2}=0,6\left(mol\right)\)

\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)

\(m_{ddHCl}=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{21,9.100\%}{8,76\%}=250\left(g\right)\)

d) \(n_{AlCl_3}=\dfrac{0,2.2}{2}=0,2\left(mol\right)\)

\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

e) \(m_{ddsau}=m_{ddHCl}+m_{Al}-m_{H_2}=250+5,4-0,3.2=254,8\left(g\right)\)

\(C\%_{AlCl_3}=\dfrac{m_{AlCl_3}}{m_{ddsau}}.100\%=\dfrac{26,7}{254,8}.100\%\approx10,48\%\)

(1). Sai

(2). Đúng

(3) Đúng

(4) Sai 

(5) Đúng

(6) Đúng

-> Chọn B

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\ Muối.tạo.thành.duy.nhất:Fe_2\left(SO_4\right)_3\\ \Rightarrow n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1\left(mol\right)\\ \Rightarrow m_{muối}=m_{Fe_2\left(SO_4\right)_3}=400.0,1=40\left(g\right)\)