Đây nhé^^

`(1)`

Gọi ct chung: \(\text{Al}_{\text{x}}\text{O}_{\text{y}}\)

`@` Theo quy tắc hóa trị: `III*x=y*II -> x/y=(II)/(III)`

`-> \text {x=2, y=3}`

`->`\(\text{CTHH: Al}_2\text{O}_3\)

\(\text{KLPT = }27\cdot2+16\cdot3=102\text{ }< \text{amu}>\)

`(2)`

Gọi ct chung: \(\text{Mg}_{\text{x}}\text{(OH)}_{\text{y}}\)

`@` Theo quy tắc hóa trị: `II*x=I*y -> x/y=I/(II)`

`-> \text {x = 1, y = 2}`

`->`\(\text{CTHH: Mg(OH)}_2\)

\(\text{KLNT = }24+\left(16+1\right)\cdot2=58\text{ }< \text{amu}>.\)

Mình lập nhanh nhé!

SO₂ hoặc SO₃

Al₂O₃

CaO

Al₂(PO₄)₃

Fe(OH)₂ hoặc Fe(OH)₃

MgSO₄

Ca₃(PO₄)₂

a)CuO

b)SO4

c)K2SO4

d)BA3(PO4)2

e)FeCl3

f)Al(NO3)3

g)PO5

h)Zn(OH)2

k)MgSO4

l)FeSO3

\(a,CTTQ:Al_x^{III}O_y^{II}\Rightarrow x\cdot III=y\cdot II\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow x=2;y=3\\ \Rightarrow Al_2O_3\\ b,CTTQ:Fe_x^{II}\left(NO_3\right)_y^I\Rightarrow x\cdot II=y\cdot I\Rightarrow\dfrac{x}{y}=\dfrac{1}{2}\Rightarrow x=1;y=2\\ \Rightarrow Fe\left(NO_3\right)_2\\ c,CTTQ:Mg_x^{II}\left(OH\right)_y^I\Rightarrow x\cdot II=y\cdot I\Rightarrow\dfrac{x}{y}=\dfrac{1}{2}\Rightarrow x=1;y=2\\ \Rightarrow Mg\left(OH\right)_2\\ d,CTTQ:N_x^{III}H_y^I\Rightarrow x\cdot III=y\cdot I\Rightarrow\dfrac{x}{y}=\dfrac{1}{3}\Rightarrow x=1;y=3\\ \Rightarrow NH_3\)

\(a,Al_2O_3\\ b,FeNO_3\\ c,Mg\left(OH\right)_2\\ d,NH_3\)

1. Na (I)

N (III)

Ca (II)

Al (III)

2. Fe2O3

NaOH

H3PO4

Mg(NO3)2

Bài 1.

a) Cu có hóa trị ll.

    O có hóa trị ll.

b) Ba có hóa trị ll.

    NO₃ có hóa trị l.

Bài 2.

a) \(BaO\Rightarrow137+16=153\left(đvC\right)\)

B) \(Al_2\left(SO_4\right)_3\Rightarrow2\cdot27+3\cdot32+16\cdot12=342\left(đvC\right)\)

Bài 3.

Theo bài ta có: \(M_M+3\cdot17=103\Rightarrow M_M=52\)

Vậy M là nguyên tử Crom.

KHHH: Cr

\(Mg_2O\)

CTHH: \(MgO\)

\(PTK_{MgO}=24+16=40\) (đvC)

\(MgO\)

\(Đặt:Al_a^{III}O_b^{II}\left(a,b:nguyên,dương\right)\\ QT.hoá.trị:a.III=II.b\Rightarrow\dfrac{a}{b}=\dfrac{II}{III}=\dfrac{2}{3}\Rightarrow a=2;b=3\\ CTTQ:Al_2O_3\\ m_{Al_2O_3}=2.27+3.16=102\left(đ.v.C\right)\)

Gọi CTHH là: \(\overset{\left(III\right)}{Al_x}\overset{\left(I\right)}{\left(OH\right)_y}\)

Ta có: \(III.x=I.y\)

\(\Leftrightarrow\dfrac{x}{y}=\dfrac{I}{III}=\dfrac{1}{3}\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)

Vậy CTHH là Al(OH)₃

\(\Rightarrow PTK_{Al\left(OH\right)_3}=27+\left(16+1\right).3=78\left(đvC\right)\)

b. Gọi CTHH là: \(\overset{\left(II\right)}{Ca_a}\overset{\left(III\right)}{\left(PO_4\right)_b}\)

Ta có: \(II.a=III.b\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{III}{II}=\dfrac{3}{2}\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\)

Vậy CTHH là Ca₃(PO₄)₂

\(\Rightarrow PTK_{Ca_3\left(PO_4\right)_2}=40.3+\left(31+16.4\right).2=310\left(đvC\right)\)