N phân tử = 1 mol phân tử

\(\Rightarrow n_{O2}=1mol;n_{N_2}=2mol;n_{CO_2}=1,5mol\)

\(\Rightarrow m_{hh}=1.32+2.28+1,5.44=154g\)

b. \(m_{hh}=0,1.56+0,2.64+0,3.65+0,25.27=44,65g\)

c. \(n_{O_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)

\(n_{HCl}=\dfrac{6,72}{22,4}=0,3mol\)

\(n_{CO_2}=\dfrac{0,56}{22,4}=0,025mol\)

\(\Rightarrow m_{hh}=0,1.32+0,05.2+0,3.36,5+0,025.44=15,35g\)

khối lượng của N phân tử oxi chính là khối lượng của 1 mol oxi, ta có: 

\(m_{O_2}=16.1=16\left(g\right)\)

tương tự với \(N_2\) và \(CO_2\)ta có:

\(m_{N_2}=28.2=56\left(g\right)\)

\(m_{CO_2}=44.1,5=66\left(g\right)\)

khối lượng của hỗn hợp là:

\(m_{hh}=m_{O_2}+m_{N_2}+m_{CO_2}=16+56+66=138\left(g\right)\)

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.

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

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`

\(a,m_{CaSO_4}=136.0,25=34\left(g\right)\\ b,n_{Cu_2O}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\\ m_{Cu_2O}=0,5.144=72\left(g\right)\\ c,n_{NH_3}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ m_{NH_3}=17.0,3=5,1\left(g\right)\\ d,m_{C_4H_{10}}=0,17.58=9,86\left(g\right)\\ e,n_{Cu\left(OH\right)_2}=\dfrac{4,5.10^{25}}{6.10^{23}}=75\left(mol\right)\\ m_{Cu\left(OH\right)_2}=98.75=7350\left(g\right)\\ g,m_{MgO}=0,48.40=19,2\left(g\right)\\ h,n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ m_{CO_2}=44.0,15=6,6\left(g\right)\\ i,m_{Al\left(OH\right)_3}=78.0,25=19,5\left(g\right)\\\)

Các câu còn lại em làm tương tự nha!

a) m_FeSO4= 0,25.152=38(g)

b) m_FeSO4= \(\dfrac{13,2.10^{23}}{6.10^{23}}.152=334,4\left(g\right)\)

c) m_NO2= \(\dfrac{8,96}{22,4}.46=18,4\left(g\right)\)

d) m_A= 27.0,22+64.0,25=21,94(g)

e) m_B= \(\dfrac{11,2}{22,4}.32+\dfrac{13,44}{22,4}.28=32,8\left(g\right)\)

g) m_C= \(64.0,25+\dfrac{15.10^{23}}{6.10^{23}}.56=156\left(g\right)\)

h) m_D= \(0,25.32+\dfrac{11,2}{22,4}.44+\dfrac{2,7.10^{23}}{6.10^{23}}.28=42,6\left(g\right)\)

hơi muộn nha<3

4.

a) \(V_{SO_2}=0.5\cdot22.4=11.2\left(l\right)\)

b) \(V_{CH_4}=\dfrac{3.2}{16}\cdot22.4=4.48\left(l\right)\)

c) \(V_{N_2}=\dfrac{0.9\cdot10^{23}}{6\cdot10^{23}}\cdot22.4=3.36\left(l\right)\)

5.

a) \(m_{Al}=0.1\cdot27=2.7\left(g\right)\)

b) \(m_{Cu\left(NO_3\right)_2}=0.3\cdot188=56.4\left(g\right)\)

c) \(m_{Na_2CO_3}=\dfrac{1.2\cdot10^{23}}{6\cdot10^{23}}\cdot106=21.2\left(g\right)\)

d) \(m_{CO_2}=\dfrac{8.96}{22.4}\cdot44=17.6\left(g\right)\)

e) \(m_K=0.5\cdot2\cdot39=39\left(g\right)\\ m_C=0.5\cdot12=6\left(g\right)\\ m_O=0.5\cdot3\cdot16=24\left(g\right)\)

a)

$n_{Al_2(SO_4)_3} = \dfrac{75,24}{27.2 + 32.3 + 16.12} = 0,22(mol)$

b)

$n_{O_2} = \dfrac{15,68}{22,4} = 0,7(mol)$
c)

$n_{H_2SO_4} = \dfrac{13,2.10^{23}}{6.10^{23}} = 2,2(mol)$

d)

$n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$n_{Al} = \dfrac{3,24}{27} = 0,12(mol)$
$n_{X} = 0,2 + 0,12 = 0,32(mol)$

e)

$n_{O_2} = \dfrac{8,94}{22,4} = 0,4(mol)$
$n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$n_Y = 0,4 + 0,1 = 0,5(mol)$

a) nAl2(SO4)3= mAl2(SO4)3/M(Al2(SO4)3)= 75,24/342=0,22(mol)

b) nO2=V(O2,đktc)/22,4=15,68/22,4=0,7(mol)

c) nH2SO4=N/6.10²³= (13,2.10²³)/(6.10²³)= 2,2(mol)

d) nX có:

Số mol Fe: nFe= mFe/M(Fe)=11,2/56=0,2(mol)

Số mol Al: nAl=mAl/M(Al)=3,24/27=0,12(mol)

e) nY có:

Số mol O2: nO2=V(O2,đktc)/22,4=8,94/22,4=447/1120(mol)

Số mol H2: nH2=V(H2,đktc)/22,4=2,24/22,4=0,1(mol