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

\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^0}2CO_2+H_2O\)

\(0.1.........0.25......0.2..........0.1\)

\(V_{C_2H_2}=2.24\left(l\right)\)

\(V_{O_2}=0.25\cdot22.4=5.6\left(l\right)\)

\(V_{CO_2}=0.2\cdot22.4=4.48\left(l\right)\)

Gọi $n_{CO} = a(mol) ; n_{CO_2} = b(mol)$

Ta có : 

$n_X = a + b = 0,25(mol)$
$m_X = 28a + 44b = 18,8.2.0,25 = 9,4(gam)$

$\Rightarrow a = 0,1 ; b = 0,15

$\%n_{CO} = \dfrac{0,1}{0,25}.100\% = 40\%$
$\%n_{CO_2} = 100\% -40\% = 60\%$

Khi nào cần z bạn?

@nguyen thi vang giúp e :))

Đáp án  C

Đặt n_CO= x mol; n C O 2 = y mol trong hỗn hợp X

→ x+ y= 22,4/22,4= 1mol (1)

CO₂+ C → t 0    2CO

z        z           2z mol

Khí Y chứa 2z+x mol khí CO và y-z mol khí CO₂

n_Y - n_X = 7,84/22,4 = 0,35 mol

→ n_Y = 0,35+ 1 = 1,35 mol

→ 2z +x + y-z = 1,35→ x+ y + z = 1,35 mol (2)

2CO₂+ Ca(OH)₂ → Ca(HCO₃)₂

Ta có: n C O 2 = 2. n C a ( H C O 3 ) 2 = 2.20,25/162 = 0,25 mol

→ y - z = 0,25 (3)

Từ (1,2,3) ta có x = 0,4; y = 0,6; z = 0,35 mol

→ % V C O = % n C O = 0 , 4 1 . 100 % = 40 %

\(n_{CO}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(Fe_2O_3+3CO\underrightarrow{^{^{t^0}}}2Fe+3CO_2\)

\(n_{CO\left(dư\right)}=x\left(mol\right),n_{CO_2}=y\left(mol\right)\)

\(\Rightarrow x+y=0.3\left(1\right)\)

\(\overline{M}=18\cdot2=36\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow28x+44y=36\cdot0.3=10.8\left(2\right)\)

\(\left(1\right),\left(2\right):x=y=0.15\)

\(m_{Fe_2O_3}=\dfrac{0.15}{3}\cdot160=8\left(g\right)\)

Đáp án C