ta có:

     Alx(OH)3=78

=>27x(16+1)3=78

=>27x=78-51

=>27x=27

=>x=1

1. 

\(PTK_{CuSO_4}=64+32+16.4=160\left(đvC\right)\)

\(PTK_{5CaCO_3}=5\left(40+12+16.3\right)=500\left(đvC\right)\)

\(PTK_{Ca\left(OH\right)_2}=40+\left(16+1\right).2=74\left(đvC\right)\)

2.

Theo đề, ta có:

\(d_{\dfrac{X}{Mg}}=\dfrac{M_X}{M_{Mg}}=\dfrac{M_X}{24}=\dfrac{4}{3}\left(lần\right)\)

=> M_X = 32(g)

Vậy X là lưu huỳnh (S)

3. 

Ta có: \(PTK_{Al_x\left(SO_4\right)_3}=27.x+\left(32+16.4\right).3=342\left(đvC\right)\)

=> x = 2

Bài 1.Phân tử khối các chất:

    \(CuSO_4\)\(\Rightarrow64+32+4\cdot16=160\left(đvC\right)\)

    \(CaCO_3\Rightarrow40+12+3\cdot16=100\left(đvC\right)\)

    \(Ca\left(OH\right)_2\Rightarrow40+16\cdot2+2=74\left(đvC\right)\)

Bài 2.Theo bài: \(\overline{M_X}=\dfrac{4}{3}\overline{M_{Mg}}=\dfrac{4}{3}\cdot24=32\left(đvC\right)\)

     Vậy X là lưu huỳnh.KHHH: S.

Bài 3. \(Al_x\left(SO_4\right)_3\) \(\Rightarrow27x+3\cdot\left(32+4\cdot16\right)=342\Leftrightarrow x=2\)

A

A

C

\(\Rightarrow27x+\left(32+16\cdot4\right)\cdot3=342\\ \Rightarrow27x+288=342\\ \Rightarrow x=2\)

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

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

\(PTK_{Al_x\left(NO_3\right)_3}=x\cdot NTK_{Al}+3NTK_N+9NTK_O=213\\ \Rightarrow27x+3\cdot14+9\cdot16=213\\ \Rightarrow27x=27\\ \Rightarrow x=1\)

x=1

Ta có: PTK = 213 

⇒ 27x + 3.(14 + 16.3) = 213

⇒ x = 1

Vậy: CTHH là Al(NO₃)₃.