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a.b.
\(n_{Fe}=\dfrac{2,8}{56}=0,05mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05 0,1 0,05 ( mol )
\(V_{H_2}=0,05.22,4=1,12l\)
\(m_{HCl}=0,1.36,5=3,65g\)
c.
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,05 0,05 ( mol )
\(m_{Cu}=0,05.64=3,2g\)
\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{65}{65}=1\left(mol\right)\\ n_{ZnCl_2}=n_{Zn}=1\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=1.136=136\left(g\right)\\ c.n_{H_2}=n_{Zn}=1\left(mol\right)\\ \Rightarrow V_{H_2}=1.22,4=22,4\left(l\right)\)
a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, \(n_{Fe_3O_4}=\dfrac{18,56}{232}=0,08\left(mol\right)\)
PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
Xét tỉ lệ: \(\dfrac{0,08}{1}>\dfrac{0,2}{4}\), ta được Fe3O4 dư.
Theo PT: \(n_{Fe}=\dfrac{3}{4}n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
a) \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,2--------------------->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
b) \(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\) => H2 hết, CuO dư
PTHH: CuO + H2 --to--> Cu + H2O
0,2<--0,2-------->0,2
=> mrắn sau pư = 24 - 0,2.80 + 0,2.64 = 20,8 (g)
c)
PTHH: RO + H2 --to--> R + H2O
0,2------>0,2
=> \(M_R=\dfrac{12,8}{0,2}=64\left(g/mol\right)\)
=> R là Cu
+) \(N_{Mg}\) = \(\dfrac{m}{M}\) = \(\dfrac{4,8}{24}\) = 0,2 mol
a) Mg + HCl -> \(MgCl_2\) + \(H_2\)
0,2 -> 0,2 (mol)
b) +) \(N_{CuO}\text{ }\)= \(\dfrac{m}{M}\) = \(\dfrac{24}{80}\) = 0,3 mol
+) \(H_2\) + CuO -> Cu + \(H_2O\)
+) Ta có: \(\dfrac{N_{H_2}}{1}\)= \(\dfrac{0,2}{1}\) < \(\dfrac{N_{CuO}}{1}\)= \(\dfrac{0,3}{1}\)
=> \(H_2\) hết. Tính toán theo \(N_{H_2}\)
+)\(H_2\) + CuO -> Cu + \(H_2O\)
Ban đầu: 0,2 0,3 0 0 }
P/ứng: 0,2 -> 0,2 -> 0,2 -> 0,2 } mol
Sau p/ư: 0 0,1 0,2 0,2 }
=> \(m_{Cu}\) = 12,8 gam .Thu được 2,8 gam Cu
Sửa đề: Thu được ZnCl2
a.b.\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,4 0,2 ( mol )
\(m_{HCl}=n.M=0,4.36,5=14,6g\)
c.\(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{32}{160}=0,2mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
\(\dfrac{0,2}{1}\) > \(\dfrac{0,2}{3}\) ( mol )
\(\dfrac{1}{15}\) 0,2 \(\dfrac{2}{15}\) 0,2 ( mol )
\(m_{Fe_2O_3\left(dư\right)}=n.M=\left(0,2-\dfrac{1}{15}\right).160=21,33g\)
\(m_{Fe}=n.M=\dfrac{2}{15}.56=7,46g\)
\(m_{H_2O}=n.M=0,2.18=3,6g\)
a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
0,1 0,2 0,1
b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(m_{HCl}=0,2\cdot36,5=7,3g\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(n_{FeO}=\dfrac{64,8}{72}=0,9\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,2------------------------->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
PTHH: FeO + H2 --to--> Fe + H2O
LTL: 0,9 > 0,2 => FeO dư
Theo pthh: nFe = nH2 = 0,2 (mol)
=> mFe = 0,2.56 =11,2 (g)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1
\(V_{H_2}=0,1.22,4=2,24l\\
m_{HCl}=\left(0,2.36,5\right).10\%=0,73g\)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\
pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\
LTL:\dfrac{0,1}{1}>\dfrac{0,1}{3}\)
=> Fe2O3 dư
\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,067\left(mol\right)\\
m_{Fe}=0,067.56=3,73g\)
a.b.\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 ( mol )
\(V_{H_2}=0,1.22,4=2,24l\)
\(m_{ddHCl}=\dfrac{0,2.36,5}{10\%}=73g\)
c.\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,1 > 0,1 ( mol )
0,1 1/15 ( mol )
\(m_{Fe}=\dfrac{1}{15}.56=3,73g\)
a,\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,2 0,4 0,2
b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
a, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
c, \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,2}{3}\), ta được Fe2O3 dư.
Theo PT: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}\left(mol\right)\Rightarrow m_{Fe}=\dfrac{2}{15}.56=\dfrac{112}{15}\left(g\right)\)