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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 )
\(m_{Fe}=n_{Fe}.M_{Fe}=0,2.56=11,2g\)
a)nO2=\(\dfrac{3.36}{22.4}\)=0,15(mol)
2KMnO4(to)→K2MnO4+MnO2+O2
Theo PT: nKMnO4=2nO2=0,3(mol)
→m=mKMnO4=0,3.158=47,4(g)
b)nH2=\(\dfrac{8.96}{22.4}\)=0,4(mol)
2H2+O2(to)→2H2O
Vì \(\dfrac{nH_2}{2}\)<nO2→O2nH2 dư
Theo PT: nH2O=nH2=0,4(mol)
→mH2O=0,4.18=7,2(g)
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 )
\(V_{H_2}=n_{H_2}.22,4=0,2.22,4=4,48l\)
a, \(2Mg+O_2\underrightarrow{^{t^o}}2MgO\)
\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
\(n_{O_2}=\dfrac{1}{2}n_{MgO}=0,025\left(mol\right)\Rightarrow V_{O_2}=0,025.22,4=0,56\left(l\right)\)
b, Có lẽ đề cho oxi tác dụng với hidro chứ không phải oxit bạn nhỉ?
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{^{t^o}}2H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{2}>\dfrac{0,025}{1}\), ta được H2 dư.
THeo PT: \(n_{H_2O}=2n_{O_2}=0,05\left(mol\right)\Rightarrow m_{H_2O}=0,05.18=0,9\left(g\right)\)
a)
n CaO = n O = 20.20%/16 = 0,25(mol)
%m CaO = 0,25.56/20 .100% = 70%
%m Ca = 100%- 70% = 30%
b)
$Ca + 2H_2O \to Ca(OH)_2 + H_2$
n H2 =n Ca = 20.30%/40 = 0,15(mol)
V = 0,15.22 4 = 3,36(lít)
n Fe3O4 = 23,2/232 = 0,1(mol)
=> n Fe(trong chất rắn) = 0,1.3 = 0,3(mol)
=> m = 0,3.56/78,9474% = 21,28(gam)
c)
$Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O$
n Fe3O4 pư = 1/4 n H2 = 0,0375(mol)
H = 0,0375/0,1 .100% = 37,5%
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,1-->0,2------>0,1-->0,1
=> VH2 = 0,1.22,4 = 2,24 (l)
mZnCl2 = 0,1.136 = 13,6 (g)
b) \(C\%_{dd.HCl}=\dfrac{0,2.36,5}{200}.100\%=3,65\%\)
c) \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,15}{1}\) => H2 hết, O2 dư
PTHH: 2H2 + O2 --to--> 2H2O
0,1--------------->0,1
=> mH2O = 0,1.18 = 1,8 (g)
a)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{H_2O}=\dfrac{1,8}{18}=0,1\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
0,1<-0,05<-------0,1
2CO + O2 --to--> 2CO2
0,2<--0,1-------->0,2
=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\V_{CO}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)
b) \(m_{CO_2}=0,2.44=8,8\left(g\right)\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{H_2O}=\dfrac{1,8}{18}=0,1mol\)
\(2CO+O_2\rightarrow2CO_2\)
a 0,5a a
\(2H_2+O_2\rightarrow2H_2O\)
0,1 0,05 \(\leftarrow\) 0,1
\(\Sigma n_{O_2}=0,5a+0,05=0,15\)
\(\Rightarrow a=n_{O_2\left(CO\right)}=0,2mol\)
\(V_{CO}=2\cdot0,2\cdot22,4=8,96l\)
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(m_{CO_2}=0,2\cdot44=8,8g\)
\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{3,36}{22,4}=0,15mol\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
0,15 0,15 ( mol )
\(m_{H_2O}=n_{H_2O}.M_{H_2O}=0,15.18=2,7g\)
2H2+O2-to>2H2O
0,15-----------------0,15
n H2=\(\dfrac{3,36}{22,4}\)=0,15 mol
=>m H2O=0,15.18=2,7g