\(n_{CO_2}=\dfrac{1.32}{44}=0.03\left(mol\right)\Rightarrow n_C=0.03\left(mol\right)\)

\(n_{H_2O}=\dfrac{0.54}{18}=0.03\left(mol\right)\Rightarrow n_H=0.03\cdot2=0.06\left(mol\right)\)

\(m_O=m_A-m_C-m_H=0.9-0.03\cdot12-0.06=0.48\left(g\right)\)

\(n_O=\dfrac{0.48}{16}=0.03\left(mol\right)\)

\(n_A=\dfrac{0.9}{180}=0.005\left(mol\right)\)

Đặt : CT : \(C_xH_yO_z\)

\(x=\dfrac{n_C}{n_A}=\dfrac{0.03}{0.005}=6\)

\(y=\dfrac{n_H}{n_A}=\dfrac{0.06}{0.005}=12\)

\(z=\dfrac{n_O}{n_A}=\dfrac{0.03}{0.005}=6\)

CT : \(C_6H_{12}O_6\)

a. 

\(n_{SO_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

Bảo toàn khối lượng : 

\(m_{O_2}=9+0.5\cdot64-17=24\left(g\right)\)

b. 

X có những nguyên tố : H và S 

c.

\(n_{H_2O}=\dfrac{9}{18}=0.5\left(mol\right)\Rightarrow n_H=0.5\cdot2=1\left(mol\right)\)

Số nguyên tử H : Số nguyên tử S = 1 : 0.5 = 2 : 1 

d.

Ta có công thức nguyên của X : \(\left(H_2S\right)_n\)

\(M_X=34n=34\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow n=1\)

\(CT:H_2S\)

C

C

\(n_{H_2S}=\dfrac{6.4}{34}=\dfrac{16}{85}\left(mol\right)\)

nH₂S= \(\dfrac{6,4}{34}=\dfrac{16}{85}\) (mol)

\(M_X=\dfrac{1.5}{\dfrac{1.775}{71}}=60\left(g\text{/}mol\right)\)

\(n_X=\dfrac{2.25}{60}=0.0375\left(mol\right)\)

CT : \(C_xH_yO_z\)

\(n_{CaCO_3}=n_{CO_2}=\dfrac{11.25}{100}=0.1125\left(mol\right)\)

\(\Rightarrow n_C=0.1125\left(mol\right)\)

\(m_{bt}=m_{CO_2}+m_{H_2O}=7.65\left(g\right)\)

\(\Rightarrow n_{H_2O}=\dfrac{7.65-0.1125\cdot44}{18}=0.15\left(mol\right)\)

\(\Rightarrow n_H=0.15\cdot2=0.3\left(mol\right)\)

Ta có : 

\(x=\dfrac{n_C}{n_X}=\dfrac{0.1125}{0.0375}=3\)

\(y=\dfrac{n_H}{n_X}=\dfrac{0.3}{0.0375}=8\)

\(M=12\cdot3+8+16z=60\left(g\text{/}mol\right)\)

\(\Rightarrow z=1\)

\(CT:C_3H_8O\)

a.

\(d_{A\text{/}O_2}=1.375\) \(\Rightarrow M_A=1.375\cdot32=44\left(g\text{/}mol\right)\)

b.

\(d_{B\text{/}kk}=0.586\) \(\Rightarrow M_B=0.586\cdot29=17\left(g\text{/}mol\right)\)

\(m_{\text{bình 1 tăng}}=m_{H_2O}=1.8\left(g\right)\)

\(\Rightarrow n_{H_2O}=\dfrac{1.8}{18}=0.1\left(mol\right)\)

\(\Rightarrow n_H=0.1\cdot2=0.2\left(mol\right)\)

\(m_{\text{bình 2 tăng}}=m_{CO_2}=3.52\left(g\right)\)

\(\Rightarrow n_{CO_2}=\dfrac{3.52}{44}=0.08\left(mol\right)\)

\(\Rightarrow n_C=0.08\left(mol\right)\)

\(m_O=m_A-m_C-m_H=1.48-0.08\cdot12-0.2=0.32\left(g\right)\)

Vậy A có 3 nguyên tố : C , H , O 

\(n_O=\dfrac{0.32}{16}=0.02\left(mol\right)\)

\(n_A=\dfrac{1.48}{74}=0.02\left(mol\right)\)

Đặt : CT là : \(C_xH_yO_z\)

\(x=\dfrac{n_C}{n_A}=\dfrac{0.08}{0.02}=4\)

\(y=\dfrac{n_H}{n_A}=\dfrac{0.2}{0.02}=10\)

\(z=\dfrac{n_O}{n_A}=\dfrac{0.02}{0.02}=1\)

CTPT của A là : \(C_4H_{10}O\)

\(yH_2+Fe_xO_y\underrightarrow{t^o}xFe+yH_2O\)

a, \(Mg\left(OH\right)_2,CuO,NaOH,\)

b, \(K_2SO_4,HNO_3,P_2O_5\)