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\(pthh:Fe_2O_3+3H_2\overset{t^o}{--->}2Fe+3H_2O\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo pt: \(n_{Fe}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)
\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
0,3 0,2 ( mol )
\(m_{Fe}=n_{Fe}.M_{Fe}=0,2.56=11,2g\)
\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{3,36}{22,4}=0,15mol\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
0,15 0,15 ( mol )
\(m_{H_2O}=n_{H_2O}.M_{H_2O}=0,15.18=2,7g\)
a) \(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
1<-----------------------------0,5
=> \(m_{KMnO_4}=1.158=158\left(g\right)\)
b) \(n_{Fe_2O_3}=\dfrac{80}{160}=0,5\left(mol\right)\)
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
0,5--->1,5
=> \(V_{H_2}=1,5.22,4=33,6\left(l\right)\)
c) \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\) => CuO dư, H2 hết
PTHH: CuO + H2 --to--> Cu + H2O
0,05<-0,05---->0,05-->0,05
=> \(n_{Cu\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
mCu = 0,05.64 = 3,2 (g)
VH2O = 0,05.22,4 = 1,12 (l)
a)\(n_{O_2}=\dfrac{11,2}{22,4}=0,5mol\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
1 0,5
\(M_{KMnO_4}=1\cdot158=158g\)
b)\(n_{Fe_2O_3}=\dfrac{80}{160}=0,5mol\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
0,5 1,5
\(V_{H_2}=1,15\cdot22,4=25,76l\)
Câu 1:
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{HCl}=\dfrac{43,8}{36,5}=1,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{1,2}{2}\) \(\Rightarrow\) HCl còn dư, Fe p/ứ hết
\(\Rightarrow n_{H_2}=0,2\left(mol\right)\) \(\Rightarrow V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)
\(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
0,3-->0,3
=> V = 0,3.22,4 = 6,72 (l)=> B
2Cu+O2-to>2CuO
0,4-----0,2-----------0,4 mol
n Cu=\(\dfrac{12,8}{64}\)=0,4 mol
=>m CuO=0,4.56=22,4g
=>Vkk=0,2.22,4.5=22,4l
a ) PTHH của phản ứng :
\(C_2H_5OH+3O_2\rightarrow2CO_2+3H_2O\)
b ) \(n_{H_2O}=\frac{5,4}{18}=0,3\) mol
Theo phản ứng trên :
\(n_{C_2H_5OH}=\frac{1}{3}n_{H_2O}=0,1\) mol \(\Rightarrow m=46.0,1=4,6\) gam
\(n_{O_2}=n_{H_2O}=0,3\) mol \(\Rightarrow V=22,4.0,3=6,72\) lít.
CuO+H2-to>Cu+H2O
0,2-----0,2
n CuO=\(\dfrac{16}{80}\)=0,2 mol
=>VH2=0,2.22,4=4,48l
\(n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{16}{80}=0,2mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,2 0,2 ( mol )
\(V_{H_2}=n_{H_2}.22,4=0,2.22,4=4,48l\)