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Pt:
Fe3O4 + 4H2SO4 → FeSO4 + Fe2(SO4)3 + 4H2O
0,1 → 0,4 0,1 0,1
Cu + Fe2(SO4)3 → CuSO4 + 2FeSO4
0,1 ←0,1 → 0,1 0,2
Rắn B là 0,1 mol Cu → x = 6,4 (g)
mdd NaOH = 62,5.1,12 = 70 (g)
=> \(n_{NaOH}=\dfrac{70.16\%}{40}=0,28\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}C_{M\left(H_2SO_4\right)}=aM\\C_{M\left(Cu\left(NO_3\right)_2\right)}=bM\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{H_2SO_4}=0,1a\left(mol\right)\\n_{Cu\left(NO_3\right)_2}=0,1b\left(mol\right)\end{matrix}\right.\)
PTHH: 2NaOH + H2SO4 --> Na2SO4 + 2H2O
0,2a<----0,1a
2NaOH + Cu(NO3)2 --> Cu(OH)2 + 2NaNO3
0,2b<-----0,1b--------->0,1b
Cu(OH)2 --to--> CuO + H2O
0,1b------------>0,1b
=> \(0,1b=\dfrac{1,6}{80}=0,02\)
=> b = 0,2
Có: nNaOH = 0,2a + 0,2b = 0,28
=> a = 1,2
Vậy \(\left\{{}\begin{matrix}C_{M\left(H_2SO_4\right)}=1,2M\\C_{M\left(Cu\left(NO_3\right)_2\right)}=0,2M\end{matrix}\right.\)
a)2NaOH+H2SO4→Na2SO4+2H2O(1)
Cu(NO3)2+2NaOH→Cu(OH)2+2NaNO3(2)
Cu(OH)2→CuO+H2O(3)
nCuO=\(\dfrac{1,6}{80}\)=0,02mol
mddNaOH=31,25×1,12=35g
nNaOH=35×16%40=0,14mol
nNaOH(2)=0,02×2=0,04mol
⇒nNaOH(1)=0,14−0,04=0,1mol
nH2SO4=0,12=0,05mol
CM(H2SO4)=\(\dfrac{0,05}{0,05}\)=1M
CM(Cu(NO3)2)=\(\dfrac{0,02}{0,05}\)=0,4M
b)nCu=\(\dfrac{2,4}{64}\)=0,0375mol
nH+=2nH2SO4=0,1mol
nNO3−=2nCu(NO3)2=0,04mol
Cu+4H++NO3−→Cu2++NO+2H2O
\(\dfrac{0,04}{1}\)>\(\dfrac{0,03751}{1}\)>\(\dfrac{0,1}{4}\)⇒ Tính theo ion H+nNO=0,14=0,025mol
⇒VNO=0,025×22,4=0,56l
a) mNaOH= 200.20%= 40(g)
=>nNaOH=1(mol)
PTHH: 2 NaOH + CuCl2 -> 2 NaCl + Cu(OH)2
Dung dịch sau khi lọc kết tủa có NaCl.
nNaCl=nNaOH= 1(mol)
nCuCl2=nCu(OH)2=nNaOH/2=1/2=0,5(mol)
mNaCl=1.58,5=58,5(g)
mCuCl2=0,5.135=67,5(g)
=> mddCuCl2=(67,5.100)/10=675(g)
mCu(OH)2=0,5.98=49(g)
=>mddNaCl=mddNaOH+ mddCuCl2 - mCu(OH)2= 200+675 - 98=777(g)
=> \(C\%ddNaCl=\dfrac{58,5}{777}.100\approx7,529\%\)
b) PTHH: Cu(OH)2 -to-> CuO + H2O
0,5__________________0,5(mol)
m(rắn)=mCuO=0,5.80=4(g)
BTKL
mX + mdd HNO3 = mdd X + mH2O + m↑
=> mdd X = 11,6 + 87,5 – 30 . 0,1 – 46 . 0,15 = 89,2g
=> C%Fe(NO3)3 = 13,565%
Bài 7 :
200ml = 0,2l
\(n_{CuCl2}=2.0,2=0,4\left(mol\right)\)
Pt : \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl|\)
1 2 1 2
0,4 0,8 0,4 0,8
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O|\)
1 1 1
0,4 0,4
a) \(n_{CuO}=\dfrac{0,4.1}{1}=0,4\left(mol\right)\)
⇒ \(m_{CuO}=0,4.40=32\left(g\right)\)
b) \(n_{NaCl}=\dfrac{0,4.2}{1}=0,8\left(mol\right)\)
⇒ \(m_{NaCl}=0,8.58,5=46,8\left(g\right)\)
\(m_{ddCuCl2}=1,35.200=270\left(g\right)\)
\(m_{ddspu}=270+100=370\left(g\right)\)
\(C_{NaCl}=\dfrac{46,8.100}{370}=12,65\)0/0
Chúc bạn học tốt
\(n_{CuSO_4}=\dfrac{15,2}{160}=0,095mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
0,095 0,19 0,095 0,095
\(m_{rắn}=m_{Cu\left(OH\right)_2}=0,095.98=9,31g\\ V_{ddNaOH}=\dfrac{0,19}{2}=0,095l\\ b)C_{M_{Na_2SO_4}}=\dfrac{0,095}{0,04+0,095}\approx0,7M\\ c)Cu\left(OH\right)_2\xrightarrow[t^0]{}CuO+H_2O\)
0,095 0,095
\(m_{rắn}=m_{CuO}=0,095.80=7,6g\)