Cho 54g dung dịch CuCl2 tác dụng vừa đủ với dung dịch KOH
a,viết phương trình hoá học của phản ứng xảy ra
b,tính khối lượng chất rắn thu được sau phản ứng
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\(n_{CuCl_2}=\dfrac{54}{135}=0,4\left(mol\right)\\ 2KOH+CuCl_2\rightarrow2KCl+Cu\left(OH\right)_2\\ n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,4\left(mol\right)\\ m_{rắn}=m_{Cu\left(OH\right)_2}=98.0,4=39,2\left(g\right)\)
M + 2HCl → MCl2 + H2↑
MO + 2HCl → MCl2 + H2O
MCl2 + 2NaOH → M(OH)2↓ + 2NaCl
M(OH)2 → MO + H2O
M + CuCl2 → MCl2 + Cu↓
\(n_{CuCl_2}=\dfrac{60,75}{135}=0,45mol\\ a)CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
0,45 0,9 0,45 0,9
\(b)m_X=m_{Cu\left(OH\right)_2}=0,45.81=36,45g\\
c)m_{ddNaOH}=\dfrac{0,9.40}{15\%}\cdot100\%=240g\\
d)m_{ddNaCl}=60,75+240-36,45=264,3g\\
C_{\%NaCl}=\dfrac{0,9.58,5}{264,3}\cdot100\%=19,92\%\\
e)n_{H_2SO_4}=\dfrac{245.20\%}{100\%.98}=0,5mol\\
H_2SO_4+Cu\left(OH\right)_2\rightarrow CuSO_4+2H_2O\\
\Rightarrow\dfrac{0,5}{1}>\dfrac{0,45}{1}\Rightarrow H_2SO_4.dư\)
\(\Rightarrow\)Dung dịch acid \(H_2SO_4\) làm tan hết chất X\(\left(Cu\left(OH\right)_2\right)\)
a) \(n_{Al}=\dfrac{32,4}{27}=1,2\left(mol\right)\)
PTHH: 2Al + 3CuCl2 --> 2AlCl3 + 3Cu
_____1,2--->1,8-------->1,2----->1,8
=> mCu = 1,8.64 = 115,2 (g)
b) \(V_{ddCuCl_2}=\dfrac{1,8}{1,5}=1,2\left(l\right)\)
c) \(AlCl_3+3NaOH\rightarrow3NaCl+Al\left(OH\right)_3\downarrow\)
\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)
a, \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
b, Ta có: \(m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Theo PT: \(n_{MgO}=n_{MgSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)
c, Ta có: m dd sau pư = 8 + 200 = 208 (g)
\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,2.120}{208}.100\%\approx11,54\%\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
\(n_{ZnCl_2}=n_{H_2}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
nMg = 4.8/24 = 0.2 (mol)
Mg + 2HCl => MgCl2 + H2
0.2.................................0.2
CuO + H2 -to-> Cu + H2O
...........0.2..........0.2
mCu = 0.2*64 = 12.8 (g)
a) PTHH: Mg + 2HCl -> MgCl2 + H2
0,2____________0,4___0,2___0,2(mol)
CuO + H2 -to-> Cu + H2O
0,2___0,2____0,2(mol)
b) =>mCu=0,2.64=12,8(g)
a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)=n_{MgCl_2}\)
\(\Rightarrow m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\)
c) Theo PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(p.ứ\right)}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl\left(p.ứ\right)}=0,3\cdot36,5=10,95\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=53,3\left(g\right)\)
\(\Rightarrow m_{ddHCl}=50\left(g\right)\) \(\Rightarrow C\%_{HCl\left(p.ứ\right)}=\dfrac{10,95}{50}\cdot100\%=21,9\%\)
a: \(CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2\downarrow+2KCl\)
b: \(n_{CuCl_2}=\dfrac{54}{135}=0.4\left(mol\right)\)
=>\(n_{Cu\left(OH\right)_2}=0.4\left(mol\right)\)
\(m_{Cu\left(OH\right)_2}=0.4\cdot\left(64+16\cdot2+2\right)=39.2\left(g\right)\)