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a)
- \(V_{CO}=n.24=0,2.24=4,8\left(l\right)\)
- \(n_{SO_3}=\dfrac{m}{M}=\dfrac{8}{80}=0,1\left(mol\right)\)
`=>` \(V_{SO_3}=n.24=0,1.24=2,4\left(l\right)\)
- \(n_{N_2}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)
`=>` \(V_{N_2}=n.24=0,5.24=12\left(l\right)\)
b)
- \(m_{Fe_2O_3}=n.M=0,25.160=40\left(g\right)\)
- \(m_{Al_2O_3}=n.M=0,15.102=15,3\left(g\right)\)
- \(n_{O_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
`=>` \(m_{O_2}=n.M=0,15.32=4,8\left(g\right)\)
c)
Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=\dfrac{m}{M}=\dfrac{8}{64}=0,125\left(mol\right)\\n_{CO_2}=\dfrac{m}{M}=\dfrac{4,4}{44}=0,1\left(mol\right)\\n_{H_2}=\dfrac{m}{M}=\dfrac{0,1}{2}=0,05\left(mol\right)\end{matrix}\right.\)
`=>` \(n_{hh}=n_{SO_2}+n_{CO_2}+n_{H_2}=0,125+0,1+0,05=0,275\left(mol\right)\)
`=>` \(V_{hh\left(\text{đ}ktc\right)}=n_{hh}.22,4=0,275.22,4=6,16\left(l\right)\)
a) mO2=20,32 gam
b) mO2=23,12 gam
c) mO2=43,04 gam
a)
Cu +1/2O2 -> CuO
2Fe + 3O2 -> Fe2O3
Ba + 1/2O2 -> BaO
-> nO2=1/2nCu +3/2nFe +1/2nBa=0,25.1/2+0,09.3/2+0,75.1/2=0,635 mol -> mO2=0,635.32=20,32 gam
b) nP=7,75/31=0,25 mol; nS=11,2/32=0,35 mol; nC=1,08/12=0,06 mol
4P + 5O2 -> 2P2O5
S + O2 -> SO2
C + O2 -> CO2
-> nO2=5/4nP + nS +nC=0,25.5/4+0,35+0,06=0,7225 mol -> mO2=0,7225.32=23,12 gam
c)
nC2H6=5,6/22,4=0,25 mol
nH2=0,896/22,4=0,04 mol
nC2H4=3,36/22,4=0,15 mol
C2H6 + 3,5O2 -> 2CO2 + 3H2O
2H2 +O2 -> 2H2O
C2H4 + 3O2 -> 2CO2 + 2H2O
-> nO2=3,5nC2H6 +1/2nH2 +3nC2H4=0,25.3,5+0,04.1/2+0,15.3=1,345 gam -> mO2=1,345.32=43,04 gam
PTHH: CuO + CO --to--> Cu + CO2
______0,1---------------------->0,1
Fe2O3 + 3CO --> 2Fe + 3CO2
_0,1----------------------->0,3
=> VCO2 = (0,1+0,3).22,4 = 8,96(l)
=> D
Fe2O3 + 2Al => Al2O3 + 2Fe
\(\frac{23}{3200}\) \(\frac{23}{1600}\)(mol)
Al0 => Al+3 + 3e
\(\frac{23}{1600}\) \(\frac{69}{1600}\)(mol)
Dung dịch X gồm :
$Mg^{2+} : 0,1$
$H^+ : 0,2$
$SO_4^{2-} : 0,075$
$NO_3^- : 0,25$
Dung dịch Y gồm :
$K^+ : 0,15$
$NH_4^+ : 0,25$
$CO_3^{2-} : 0,15$
$Cl^- : 0,1$
Quá trình oxi hóa: \(\overset{+3}{Fe_2}+6e\rightarrow\overset{0}{2Fe}\)
0,25 -->1,5 (mol)
\(\Rightarrow\) Chọn D
bn cho mình hỏi sao 0,25 thành 1,5 (mol) dc vậy???