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PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\\n_{H_2SO_4}=\dfrac{784\cdot10\%}{98}=0,8\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,6}{1}< \dfrac{0,8}{1}\) \(\Rightarrow\) H2SO4 còn dư, Fe phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{FeSO_4}=n_{H_2}=0,6mol\\n_{H_2SO_4\left(dư\right)}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,6\cdot152=91,2\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,2\cdot98=19,6\left(g\right)\\m_{H_2}=0,6\cdot2=1,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=816,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{91,2}{816,4}\cdot100\%\approx11,17\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{19,6}{816,4}\cdot100\%\approx2,4\%\end{matrix}\right.\)
\(n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\)
\(n_{H2SO4}=\dfrac{784.10\%}{98}=0,8\left(mol\right)\)
PTHH : \(Fe+H_2SO_4-->FeSO_4+H_2\uparrow\)
Theo pthh : \(n_{H2}=n_{FeSO4}=n_{H2SO4\left(pứ\right)}=n_{Fe}=0,6\left(mol\right)\)
\(\Rightarrow n_{H2SO4\left(dư\right)}=0,8-0,6=0,2\left(mol\right)\)
Áp dụng ĐLBTKL :
mFe + m(dd H2SO4) = m(ddspu) + mH2
=> 33,6 + 784 = m(ddspu) + 0,6.2
=> m(ddspu) = 816,4(g)
\(\Rightarrow\left\{{}\begin{matrix}C\%FeSO_{\text{4}}=\dfrac{0,6.152}{816,4}\cdot100\%\approx11,17\%\\C\%H_2SO_{4\left(dư\right)}=\dfrac{0,2.98}{816,4}\cdot100\%\approx2,4\%\end{matrix}\right.\)
,nFe2(SO4)3=0.25*1=0.25mol=>nFe3+=0.5 mol
spư thu dc ran z ma ko tan trong H2SO4 loang=>ran Z la Cu,nCu dư=3.28/64=0.05125mol=>Fe tan het
va toan bo Fe(3+) chuyen len het Fe(2+) do Cu dư
dat nFe=x,nCu pư=y=>56x+64y+3.28=17.8 (1)
mat khac theo bt e
Fe=Fe(2+)+2e................Fe(3+)+1e=F...
x-------------> 2x 0.5---->0.5
Cu=Cu(2+)+2e
y------------->2y
vi ne cho=ne nhan=>2x+2y=0.5(2)=>x=0.185,y=0.065
mCu=3.28+0.065*64=7.44g
b) \(Al\underrightarrow{1}Al_2O_3\underrightarrow{2}Al\underrightarrow{3}Al_2\left(SO_4\right)_3\underrightarrow{4}Al\left(OH\right)_3\underrightarrow{5}Al_2O_3\)
(1) \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
(2) \(2Al_2O_3\xrightarrow[điện.phân.nóng.chảy]{criolit}4Al+3O_2\)
(3) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
(4) \(Al_2\left(SO_4\right)_3+6KOH\rightarrow2Al\left(OH\right)_3+3K_2SO_4\)
(5) \(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)
Chúc bạn học tốt
Câu 3 :
200ml = 0,2l
\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,03 0,03 0,03
b) \(n_{H2}=\dfrac{0,03.1}{1}=0,03\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,03.24,79=0,7437\left(l\right)\)
c) \(n_{ZnCl2}=\dfrac{0,03.1}{1}=0,03\left(mol\right)\)
\(C_{M_{ZnCl2}}=\dfrac{0,03}{0,2}=0,15\left(M\right)\)
Chúc bạn học tốt
Ta có PTHH
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
\(m_{H_2SO_4}=\frac{14,6\times150}{100}=21,9\left(g\right)\Rightarrow n_{H_2SO_4}=\frac{219}{980}\left(mol\right)\)
\(n_{Al}=\frac{2}{3}n_{H_2SO_4}=\frac{73}{490}\left(mol\right)\Rightarrow m_{Al}=\frac{73}{490}\times27=\frac{1971}{490}\left(g\right)\)
\(m_{HCl}=\frac{9,8\times150}{100}=14,7\left(g\right)\Rightarrow n_{HCl}=\frac{147}{365}\left(mol\right)\)
\(\Rightarrow n_{Al}=\frac{1}{3}n_{HCl}=\frac{49}{365}\left(mol\right)\Rightarrow m_{Al}=\frac{49}{365}\times27=\frac{1323}{365}\left(g\right)\)
\(\Rightarrow m=m_{Al\left(1\right)}+m_{Al\left(2\right)}=\frac{1971}{490}+\frac{1323}{365}\approx7,65\left(g\right)\)
Có
\(n_{Al_2\left(SO_4\right)_3}=\frac{1}{3}n_{H_2SO_4}=\frac{73}{980}\left(mol\right)\Rightarrow C\%\left(Al_2\left(SO_4\right)_3\right)=\frac{\frac{73}{980}\times342}{150+7,65}.100\%\approx4,04\%\)
\(n_{AlCl_3}=\frac{1}{3}n_{HCl}=\frac{49}{635}\left(mol\right)\Rightarrow C\%\left(AlCl_3\right)=\frac{\frac{49}{635}}{150+7,65}.100\%\approx6,53\%\)
\(a,n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow n_{HCl}=0,4\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,4\cdot36,5=14,6\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{14,6}{200}\cdot100\%=7,3\%\\ b,n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V_{H_2\left(đkc\right)}=0,2\cdot24,79=4,958\left(l\right)\\ c,m_{H_2}=0,2\cdot2=0,4\left(g\right)\\ n_{FeCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{CT_{FeCl_2}}=0,2\cdot127=25,4\left(g\right)\\ \Rightarrow m_{dd_{FeCl_2}}=11,2+200-0,4=210,8\left(g\right)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{25,4}{210,8}\cdot100\%\approx12,05\%\)
a) Vì: mA < 400 (g) nên phải có khí thoát ra → muối có dạng MHSO4 và khí là: CO2
b)
c) Tác dụng được với: MgCO3, Ba(HSO3)2, Al2O3, Fe(OH)2, Fe, Fe(NO3)2
Pt: 2NaHSO4 + MgCO3 → Na2SO4 + MgSO4 + CO2↑ + H2O
2NaHSO4 + Ba(HSO3)2 → BaSO4 + Na2SO4 + SO2↑ + 2H2O
6NaHSO4 + Al2O3 → 3Na2SO4 + Al2(SO4)3 + 3H2O
2NaHSO4 + Fe(OH)2 → Na2SO4 + FeSO4 + 2H2O
2NaHSO4 + Fe → Na2SO4 + FeSO4 + H2↑
12NaHSO4 + 9Fe(NO3)2 → 5Fe(NO3)3 + 2Fe2(SO4)3 + 6Na2SO4 + 3NO↑ + 6H2O
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(n_{H_2SO_4\left(bđ\right)}=\dfrac{19,6\%.200}{98}=0,4\left(mol\right)\)
\(n_{Fe}=n_{H_2SO_4\left(pứ\right)}=0,2\left(mol\right)\)
=> \(n_{H_2SO_4\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)
\(m_{ddsaupu}=11,2+200-0,2.2=210,8\left(g\right)\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2.98}{210,8}.100=9,3\%\)
PTHH: Fe+H2SO4→H2+FeSO4
nFe=\(\dfrac{11,2}{56}\)=0,2(mol)
mH2SO4=\(\dfrac{19,6\%.200}{100\%}\)=39,2(g)
nH2SO4=\(\dfrac{39,2}{98}\)=0,4(mol)
Vì Fe đã pứng hết,theo PTHH ta có:nFe=nH2SO4=0,2(mol)
⇒nH2SO4 dư=0,4-0,2=0,2(mol)
⇒mH2SO4 dư=0,2.98=19,6(g)
m dung dịch sau pứng:200+11.2=211,2(g)
⇒C%H2SO4 sau pứng=\(\dfrac{19,6.100\%}{211,2}\)(xấp xỉ) 9,3%