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PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
Ta có: \(n_{CuCl_2}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}\\n_{NaOH}=0,4\left(mol\right)=n_{NaCl}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\m_{CuO}=0,2\cdot80=16\left(g\right)\\m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\\C\%_{NaOH}=\dfrac{0,4\cdot40}{200}\cdot100\%=8\%\end{matrix}\right.\)
\(n_{CuCl_2}=0,3.0,5=0,15\left(mol\right)\)
PT: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)
Theo PT: \(n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,15\left(mol\right)\Rightarrow m_{Cu\left(OH\right)_2}=0,15.98=14,7\left(g\right)\)
\(n_{NaOH}=2n_{CuCl_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,3}{0,25}=1,2\left(M\right)\)
Bài 8: Bạn bổ sung thêm đề phần này nhé.
Bài 9: Bài này giống bài 2 bên dưới nhé.
Bài 10:
\(n_{Fe\left(NO_3\right)_3}=0,3.1=0,3\left(mol\right)\)
PT: \(Fe\left(NO_3\right)_3+3NaOH\rightarrow3NaNO_3+Fe\left(OH\right)_3\)
a, \(n_{NaOH}=3n_{Fe\left(NO_3\right)_3}=0,9\left(mol\right)\Rightarrow V_{NaOH}=\dfrac{0,9}{2}=0,45\left(l\right)\)
b, \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{Fe\left(NO_3\right)_3}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe_2O_3}=0,15.160=24\left(g\right)\)
Bài 11:
Ta có: \(n_{NaOH}=\dfrac{200.12\%}{40}=0,6\left(mol\right)\)
PT: \(2NaOH+FeCl_2\rightarrow2NaCl+Fe\left(OH\right)_2\)
a, \(n_{FeCl_2}=n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,3\left(mol\right)\Rightarrow C\%_{FeCl_2}=\dfrac{0,3.127}{100}.100\%=38,1\%\)
b, \(n_{NaCl}=n_{NaOH}=0,6\left(mol\right)\)
Ta có: m dd sau pư = 200 + 100 - 0,3.90 = 273 (g)
\(\Rightarrow C\%_{NaCl}=\dfrac{0,6.58,5}{273}.100\%\approx12,86\%\)
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
a.CuCl2 + 2NaOH -> Cu(OH)2 + 2NaCl
0.15 0.3 0.15 0.3
Cu(OH)2 -> CuO + H2O
0.15 0.15
nNaOH = 0.3 mol
\(CM_{CuCl2}=\dfrac{0.15}{2}=0.075M\)
b.Vdd sau phản ứng = 0.2 + 0.15 = 0.35l
\(CM_{NaCl}=\dfrac{0.3}{0.35}=0.86M\)
c.mCuO = \(0.15\times80=12g\)
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)
\(n_{NaOH}=\dfrac{200.10\%}{40}=0,5\left(mol\right)\)
\(PTHH:CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)
bđ: 0,3 0,5
pứ: 0,25 0,5 0,5 0,25
[ ]: 0,05 0 0,5 0,25
\(PTHH:Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
(mol) 0,25 0,25
\(a.m_C=80.0,25=20\left(g\right)\)
\(b.m_{NaCl}=58,5.0,5=29,25\left(g\right)\\ m_{Cu\left(OH\right)_2}=0,25.98=24,5\left(g\right)\\ m_{CuCl_2\left(du\right)}=135.0,05=6,75\left(g\right)\)
\(c.m_{ddspu}=100+200-24,5=275,5\left(g\right)\\ C\%_{ddCuCl_2\left(du\right)}=\dfrac{135.0,05}{275,5}.100=2,45\left(\%\right)\\ C\%_{ddNaCl}=\dfrac{0,5.58,5}{275,25}.100=10,62\left(\%\right)\)
\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)
Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)
Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)
\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)
\(n_{K2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)
Pt : \(K_2O+H_2O\rightarrow2KOH|\)
1 1 2
0,4 0,8
a) \(n_{KOH}=\dfrac{0,4.2}{1}=0,8\left(mol\right)\)
⇒ \(m_{KOH}=0,8.56=44,8\left(g\right)\)
\(C_{KOH}=\dfrac{44,8.100}{500}=8,96\)0/0
b) Pt : \(2KOH+CuSO_4\rightarrow K_2SO_4+Cu\left(OH\right)_2|\)
2 1 1 1
0,4 0,2
\(n_{Cu\left(OH\right)2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
⇒ \(m_{Cu\left(OH\right)2}=0,2.98=19,6\left(g\right)\)
Chúc bạn học tốt
Mình xin lỗi bạn nhé , bạn sửa lại giúp mình câu b)
250ml = 0,25l
\(n_{CuCl2}=2.0,25=0,5\left(mol\right)\)
Pt : \(2KOH+CuCl_2\rightarrow2KCl+Cu\left(OH\right)_2|\)
2 1 2 1
0,4 0,5
Lập tỉ số so sánh : \(\dfrac{0,4}{2}< \dfrac{0,25}{1}\)
⇒ KOH phản ứng hết , CuCl2 dư
⇒ Tính toán dựa vào số mol của KOH
Nhưng kết quả vẫn đúng bạn nhé
\(2NaOH+CuCl_2\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
\(m_{NaOH}=\frac{40.35}{100}=14\Rightarrow n_{NaOH}\frac{14}{40}=0.35\)
a,Theo pt \(n_{CuCl_2}=\frac{1}{2}n_{NaOH}=\frac{1}{2}.0.35=0.175\left(mol\right)\Rightarrow V_{CuCl_2}=\frac{0.175}{2}=0.0875\left(l\right)\)
b,theo pt:\(n_{NaCl}=n_{NaOH}=0.35,n_{Cu\left(OH\right)_2}=\frac{1}{2}n_{NaOH}=0.175\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0.175.98=17.15\left(g\right)\)
\(\Rightarrow m_{NaCl}=0.35.58.5=20.475\left(g\right)\)