a) \(M_{NH_4}=14+1.4=18\left(DvC\right)\\ \%N=\dfrac{14}{18}.100\%=78\%\\ \%H=\dfrac{1.4}{18}.100\%=22\%\)

\(M_{OH}=16+1=17\left(DvC\right)\\ \%O=\dfrac{16}{17}.100\%=94\%\\ \%H=100\%-94\%=6\%\)

`@` `\text {dnammv}`

\(\text{Al}_2\text{O}_3:\)

`- \text {NTK:}`

\(+\text{Al: 27 amu}\)

\(+\text{O: 16 amu}\)

`->`\(\text{PTK}_{\text{Al}_2\text{O}_3}=27\cdot2+16\cdot3=102\text{ }< \text{amu}>\)

\(\text{Cu(OH)}_2:\)

`- \text {NTK:}`

\(+\text{Cu: 64 amu}\)

\(+\text{H: 1 amu}\)

\(+\text{O: 16 amu}\)

`->`\(\text{PTK}_{\text{Cu}\left(\text{OH}\right)_2}=64+\left(16+1\right)\cdot2=98\text{ }< \text{amu}>.\)

1. Ta có: 56 x 2 + ( 32 + 16 x 4 )x = 400

=> x = 3

=> CTHH: Fe₂(SO4)₃

2. Cách giải tương tự 1

=> CTHH: Al₂(SO4)₃

3. Cách giải tương tự 1

=> CTHH: K₂SO₄

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

Các câu còn lại em làm tương tự nha!

a, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{42,75}{342}=0,125\left(mol\right)\)

\(n_O=12n_{Al_2\left(SO_4\right)_3}=1,5\left(mol\right)\)

b, \(n_{O\left(Al_2O_3\right)}=3n_{Al_2O_3}=2n_{O\left(Al_2\left(SO_4\right)_3\right)}=3\left(mol\right)\)

\(\Rightarrow n_{Al_2O_3}=1\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=102\left(g\right)\)

M(OH)₂ + H₂SO₄-----> MSO₄ + 2H₂O

d= MSO₄/M(OH)₂=\(\frac{M+96}{M+34}=2,069\)

1,069M=96-34.2,069

M=24=Mg

a)

$x = 2 ; y = 3$

PTHH : $2Al(OH)_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 6H_2O$

b)

Tỉ lệ số phân tử $Al(OH)_3$ : số phân tử $H_2SO_4$ là 2 : 3

c)

$\%Al = \dfrac{27}{78}.100\% =34,6\%$

d)

$n_{Al(OH)_3} = \dfrac{7,8}{78} = 0,1(mol)$

Theo PTHH : $n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al(OH)_3} = 0,05(mol)$

$m_{Al_2(SO_4)_3} = 0,05.342 = 17,1(gam)$