\(n_{H_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)

gọi x la so mol cua Al

y la so mol cua Mg

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

de: x \(\rightarrow\) 1,5x

Mg + 2HCl \(\rightarrow\) MgCl₂ + H₂

de: y \(\rightarrow\) y

Ta co: 27x + 24y = 1,56

1,5x + y = 0,08

=> x = 0,04 y = 0,02

\(m_{Al}=0,04.27=1,08g\)

\(\%m_{Al}=\dfrac{1,08}{1,56}.100\%\approx69,23\%\)

\(\%m_{Mg}=100-69,23=30,77\%\)

CT: R₂O_x

R₂O_x + 2xHCl \(\rightarrow\) 2RCl_x + xH₂O

pt: 2R + 16x 2R + 71x

de: 5,6 11,1

Ta co: \(11,1\left(2R+16x\right)=5,6\left(2R+71x\right)\)

\(\Leftrightarrow22,2R+177,6x=11,2R+397,6x\)

\(\Leftrightarrow11R=220x\)

\(\Leftrightarrow R=20x\)

biện luận:

+ x = 1 => R = 20(loại)

+ x = 2 => R = 40 (lay)

=> CT: CaO

CT: H_xO_y

Ta co: \(\dfrac{x}{16y}=\dfrac{1}{8}\Rightarrow\dfrac{x}{y}=\dfrac{16}{8}=\dfrac{2}{1}\)

=> CT: H₂O

CaCO₃ \(\underrightarrow{t^o}\) CaO + CO₂

Ta co: \(m_{CaCO_3}=m_{CaO}+m_{CO_2}\)

\(\Rightarrow m_{CaCO_3}=28+22=50g\)

\(\%CaCO_3=\dfrac{50}{60}.100\%\approx83,33\%\)

\(n_{H_2}=\dfrac{1,2}{2}=0,6\left(mol\right)\)

so phtu có trong 1,2g H2:

0,6.6.10²³ = 3,6.10²³

gọi x la so g Mg

\(n_{Mg}=\dfrac{x}{24}\left(mol\right)\)

so phtu Mg: \(\dfrac{x}{24}.6.10^{23}=3,6.10^{23}\)

\(\Rightarrow6x.10^{23}=86,4.10^{23}\)

\(\Rightarrow x=14,4g\)

x-28=1

=>x=28+1

=>x=29

H₂O

1, \(m_{H_2SO_4}=\dfrac{294.20}{100}=58,8g\)

CT: R₂O_x

R₂O_x + xH₂SO₄ \(\rightarrow\) R₂(SO₄)_x + xH₂O

pt: 2R+16x 98x

de: 30,4 58,8

Ta co: 58,8( 2R + 16x ) = 2979,2x

=> 117,6R + 940,8x = 2979,2x

=> 117,6R = 2038,4x

=> \(R=\dfrac{52}{3}x\)

biện luận:

+ x = 1 => R = 52/3 (loai)

+ x= 2 => R = 104/3 (loai)

+ x = 3 => R = 52 (lay)

=> CT: Cr₂O₃