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Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)
PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
____0,5____________0,5 (mol)
a, \(m_{CuCl_2}=0,5.135=67,5\left(g\right)\)
b, Có: m dd sau pư = mCuO + m dd HCl = 40 + 200 = 240 (g)
\(\Rightarrow C\%_{CuCl_2}=\dfrac{67,5}{240}.100\%=28,125\%\)
Bạn tham khảo nhé!
\(n_{FeO}=\dfrac{3.2}{72}=\dfrac{2}{45}\left(mol\right)\)
\(FeO+H_2\underrightarrow{^{^{t^0}}}Fe+H_2O\)
\(\dfrac{2}{45}....\dfrac{2}{45}....\dfrac{2}{45}\)
\(V_{H_2}=\dfrac{2}{45}\cdot22.4=1\left(l\right)\)
\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{^{t^0}}FeCl_3\)
\(\dfrac{2}{45}.............\dfrac{2}{45}\)
\(m_{FeCl_3}=\dfrac{2}{45}\cdot162.5=7.22\left(g\right)\)
\(n_{HCl}=\dfrac{80\cdot14.6\%}{36.5}=0.32\left(mol\right)\)
\(n_{FeO}=\dfrac{7.2}{72}=0.1\left(mol\right)\)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\)
Lập tỉ lệ :
\(\dfrac{0.1}{1}< \dfrac{0.32}{2}\Rightarrow HCldư\)
\(m_{HCl\left(dư\right)}=\left(0.32-0.1\cdot2\right)\cdot36.5=4.38\left(g\right)\)
\(m_{FeCl_2}=0.1\cdot127=12.7\left(g\right)\)
\(m_{HCl}=\dfrac{14,6.80}{100}=11,68\left(g\right)\Rightarrow n_{HCl}=\dfrac{11,68}{36,5}=0,32\left(mol\right)\);
\(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)
\(PTHH:FeO+2HCl\rightarrow FeCl_2+H_2O\)
Mol: 0,1 0,2 0,1
Ta có tỉ lệ:\(\dfrac{0,32}{2}>\dfrac{0,1}{1}\) =>HCl dư,FeO phản ứng hết
a)mHCl dư=(0,32-0,2).36,5=4,38 (g)
b)\(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
Bài 3:
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
a, PTHH: Fe3O4 + 4H2 --to--> 3Fe + 4H2O
0,1<------0,4
Zn + 2HCl ---> ZnCl2 + H2
0,4<-------------------------0,4
b, mFe3O4 = 0,1.232 = 23,2 (g)
c, mZn = 0,4.65 = 26 (g)
Bài 4:
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
a, PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,1---->0,2---------------->0,1
b, VH2 = 0,1.22,4 = 2,24 (l)
c, \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)
\(n_{HCl}=0.2\cdot1=0.2\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(........0.2..............0.1\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(n_{CuO}=\dfrac{16}{80}=0.2\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(0.1.......0.1....0.1\)
\(\Rightarrow CuOdư\)
\(m_{Cu}=0.1\cdot64=6.4\left(g\right)\)
Sửa đề : 11.2 g sắt
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(0.2....0.2.................0.2\)
\(m_{FeSO_4}=0.2\cdot152=30.4\left(g\right)\)
\(C_{M_{H_2SO_4}}=\dfrac{0.2}{0.05}=4\left(M\right)\)
\(n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\\ FeO+2HCl\rightarrow FeCl_2+H_2O\\ n_{FeO}=n_{FeCl_2}=n_{H_2O}=\dfrac{0,8}{2}=0,4\left(mol\right)\\ a,m_{FeO}=72.0,4=28,8\left(g\right)\\ b,C1:m_{sp}=m_{FeO}+m_{HCl}=28,8+29,2=58\left(g\right)\\ C2:m_{sp}=m_{FeCl_2}+m_{H_2O}=127.0,4+18.0,4=58\left(g\right)\)
\(a.n_{HCl}=0,8\left(mol\right)\\ FeO+2HCl\rightarrow FeCl_2+H_2O\\ n_{FeO}=\dfrac{1}{2}n_{HCl}=0,4\left(mol\right)\\ m_{FeO}=0,4.72=28,8\left(g\right)\\ b.n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,4\left(mol\right)\\ m_{FeCl_2}=0,4.127=50,8\left(g\right)\\ n_{H_2O}=\dfrac{1}{2}n_{HCl}=0,4\left(mol\right)\\ \Rightarrow m_{H_2O}=0,4.18=7,2\left(g\right)\)
a) nFe=0,1(mol); nHCl=0,4(mol)
PTHH: Fe + 2 HCl -> FeCl2 + H2
Ta có: 0,1/1 < 0,4/2
=> Fe hết, HCl dư, tish theo nFe.
b) nH2=nFeCl2=Fe=0,1(mol)
=> V(H2,đktc)=0,1.22,4=2,24(l)
c) mFeCl2=127.0,1=12,7(g)
a) nFe=0,1(mol); nHCl=0,4(mol) PTHH: Fe + 2 HCl -> FeCl2 + H2 Ta có: 0,1/1 < 0,4/2 => Fe hết, HCl dư, tish theo nFe. b) nH2=nFeCl2=Fe=0,1(mol) => V(H2,đktc)=0,1.22,4=2,24(l) c) mFeCl2=127.0,1=12,7(g)
Sửa đề: tác dụng với HCl
\(n_{FeO}=\dfrac{7.2}{72}=0.1\left(mol\right)\)
a: \(FeO+2HCl\rightarrow FeCl_2+H_2O\)
0,1 0,1
b: \(m_{FeCl_2}=0.1\left(56+35.5\cdot2\right)=12.7\left(g\right)\)
\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{7,2}{72}=0,1mol\)
PTHH: FeO + 2HCl \(\rightarrow\) FeCl2 + H2
TL: 1 : 2 : 1 : 1
Mol: 0,1 \(\rightarrow\) 0,1
b. \(m_{FeCl_2}=n_{FeCl_2}.M_{FeCl_2}=0,1.127=12,7g\)