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

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

\(N_2+3H_2\leftrightarrow2NH_3\\ n_{NH_3}=a\left(mol\right)\\ n_{sau}=14=9-\dfrac{1}{2}a+6-\dfrac{3}{2}a+a\\ a=1\\ n_{N_2}:1>n_{H_2}:3\\ H=\dfrac{1}{\dfrac{2}{3}.6}.100\%=25\%\)

A/ap dung quy tac duong cheo la ra roi

M hh=19.2=38

NO    30                8

                  38

No2   46               8

Vay NO/NO2=1

VNO=VNO2=3,36 lit

Cac cau con lai tuong tu vay

a, m_X = 0,2.32 + 0,15.28 = 10,6 (g)

n_X = 0,2 + 0,15 = 0,35 (mol)

=> M_X = \(\dfrac{10,6}{0,35}=30,3\left(\dfrac{g}{mol}\right)\)

=> d_X/kk = \(\dfrac{30,3}{29}=1,05\)

b, m_Y = 0,5.44 + 2.2 = 26 (g)

n_Y = 0,5 + 2 = 2,5 (mol)

=> M_Y = \(\dfrac{26}{2,5}=10,4\left(\dfrac{g}{mol}\right)\)

=> d_Y/O2 = \(\dfrac{10,4}{32}=0,325\)

c, m_A = 17,75 + 8,4 = 26,15 (g)

n_A = \(\dfrac{17,75}{71}+\dfrac{8,4}{28}=0,55\left(mol\right)\)

=> M_A = \(\dfrac{26,15}{0,55}=47,6\left(\dfrac{g}{mol}\right)\)

=> d_A/CO2 = \(\dfrac{47,6}{44}=1,1\)

Mình làm mẫu 3 ý đầu rồi mấy ý sau bạn tự làm nhé

bạn là chó đúng ko?

\(a.\)

\(GS:\)

\(n_{hh}=1\left(mol\right)\)

\(Đặt:n_{N_2}=a\left(mol\right),n_{CO_2}=b\left(mol\right)\)

\(\Rightarrow a+b=1\left(1\right)\)

\(m_A=28a+44b=18\cdot2=36\left(2\right)\)

\(\left(1\right),\left(2\right):\)

\(a=b=0.5\)

\(\%m_{N_2}=\dfrac{0.5\cdot28}{0.5\cdot28+0.5\cdot44}\cdot100\%=38.89\%\)

\(\%m_{CO_2}=61.11\%\)

\(b.\)

\(\dfrac{n_{N_2}}{n_{CO_2}}=\dfrac{0.5}{0.5}=\dfrac{1}{1}\)

\(n_{N_2}=n_{CO_2}=\dfrac{1}{2}\cdot n_A=\dfrac{0.2}{2}=0.1\left(mol\right)\)

\(Đặt:n_{CO_2}=x\left(mol\right)\)

\(\overline{M}=\dfrac{0.1\cdot28+0.1\cdot44+44x}{0.2+x}=20\cdot2=40\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow x=0.2\)

\(m_{CO_2\left(cầnthêm\right)}=0.2\cdot44=8.8\left(g\right)\)