Câu `5`:

`V_(CO2) = n . 22,4 = 0,1 . 22,4 =2,24 ` (l)

`V_(H_2) = n.22,4 = 0,2 . 22,4=4,48 `( l)

`V_(O_2) = n . 22,4 = 0,7 . 22,4 =15,68` (l) 

`=> V_X= 2,24 + 4,48 + 15,68 = 22,4`(l)

`->`Chọn `C`

Câu `6: A `

Câu `7`:

Cân bằng PT: `Fe_2O_3 + 6HCl  -> 2FeCl_3 + 3H_2O`

`n_(Fe_2O_3)= 8/(2.56 + 3.16) = 0,05` (mol)

`n_(HCl) = ( 0,05 .6)/1  = 0,3 ` (mol)

`m_(HCl) = 0,3 . (1 + 35,5) = 10,95` (g)

`->` Chọn `D`

Câu `8`:

Nguyên tử khối của oxi `= 12 : 3/4 =16` ( đvC) 

`->` Chọn `C`

Câu `9`: `A`

Câu `11`: `=40+ 2( 2.1 + 31 + 4.16) =234` (g)

`->` Chọn `A`

Câu `12`:`C`

a, m_CaO = 0,5.56 = 28 (g)

b, \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

c, \(n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)

d, \(V_{hhk}=0,2.22,4+0,3.22,4=11,2\left(l\right)\)

e, \(\%m_{Cu}=\dfrac{64}{64+32+16.4}.100\%=40\%\)

Bạn tham khảo nhé!

a) mCaO=nCaO.M(CaO)=0,5.56=28(g)

b) nCO2=V(CO2,dktc)=6,72/22.4=0,3(mol)

c) nH2SO4=mH2SO4/M(H2SO4)=24,5/98=0,25(mol)

d) V(hh H2,NH3)=(0,3+0,2).22,4=11,2(l)

e) %mCu/CuSO4=(64/160).100=40%

Chúc em học tốt!

Câu 1:

a) \(m_{Fe_2\left(SO_4\right)_3}=0,15.400=60\left(g\right)\)

b) \(m_{MgCl_2}=0,05.95=4,75\left(g\right)\)

c) \(m_{H_2}=0,2.2=0,4\left(g\right)\)

d) \(n_{N_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\Rightarrow m_{N_2}=0,2.28=5,6\left(g\right)\)

e) \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\Rightarrow m_{O_2}=0,3.32=9,6\left(g\right)\)

Câu 2:

a) \(V_{NO_2}=0,25.22,4=5,6\left(mol\right)\)

b) \(V_{CO_2}=0,3.22,4=6,72\left(mol\right)\)

c) \(n_{Cl_2}=\dfrac{3,55}{35,5}=0,1\left(mol\right)\Rightarrow V_{Cl_2}=0,1.22,4=2,24\left(l\right)\)

d) \(n_{N_2O}=\dfrac{1,32}{44}=0,03\left(mol\right)\Rightarrow V_{N_2O}=0,03.22,4=0,672\left(l\right)\)

\(n_{CO_2}=0,44:44=0,01\left(mol\right)\\ n_{H_2}=0,02:2=0,01\left(mol\right)\\ n_{hh}=0,01+0,01=0,02\left(mol\right)\)

Vì C có 0,02 mol nên => Chọn câu C

\(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{0,44}{44}=0,01\left(mol\right)\\n_{H_2}=\dfrac{0,02}{2}=0,01\left(mol\right)\\V_{CO_2}=0,01.22,4=0,224\left(lít\right)\\V_{H_2}=0,01.22,4=0,224\left(lít\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{hh}=0,01+0,01=0,02\left(mol\right)\\V_{hh}=0,224+0,224=0,448\left(lít\right)\end{matrix}\right.\)

Mik thấy đề sai, nếu sai thì mik nghĩ là C nha

a.

\(m_{Al}=0.5\cdot27=13.5\left(g\right)\)

\(m_{CO_2}=\dfrac{6.72}{22.4}\cdot44=13.2\left(g\right)\)

\(m_{N_2}=\dfrac{5.6}{22.4}\cdot28=7\left(g\right)\)

\(m_{CaCO_3}=0.25\cdot100=25\left(g\right)\)

b.

\(m_{hh}=\dfrac{3.36}{22.4}\cdot2+\dfrac{5.6}{22.4}\cdot28+0.2\cdot44=16.1\left(g\right)\)

PTHH: CuO + CO --t^o--> Cu + CO₂

______0,1---------------------->0,1

Fe₂O₃ + 3CO --> 2Fe + 3CO₂

_0,1----------------------->0,3

=> V_CO2 = (0,1+0,3).22,4 = 8,96(l)

=> D

Bài 5:

\(m_{Y}=m_{SO_2}+m_{CH_4}=\dfrac{3,36}{22,4}.64+\dfrac{13,44}{22,4}.16=19,2(g)\)

Bài 6:

\(V_{CO_2}=0,15.22,4=3,36(l)\\ V_{NO_2}=0,2.22,4=4,48(l)\\ V_{SO_2}=0,02.22,4=0,448(l)\\ V_{N_2}=0,03.22,4=0,672(l)\)

\(\left\{{}\begin{matrix}n_{C_nH_{2n-2}}+n_{H_2}=\dfrac{8,96}{22,4}=0,4\\n_{C_nH_{2n-2}}=n_{H_2}\end{matrix}\right.\)

=> \(n_{C_nH_{2n-2}}=n_{H_2}=0,2\left(mol\right)\)

\(n_{H_2\left(pư\right)}=0,4-\dfrac{6,72}{22,4}=0,1\left(mol\right)\)

Bảo toàn liên kết: \(2.n_{C_nH_{2n-2}}=n_{H_2\left(pư\right)}+n_{Br_2}\)

=> \(n_{Br_2}=0,3\left(mol\right)\)

=> B