\(n_{H_2SO_4}=5a\left(mol\right),n_{HCl}=3a\left(mol\right)\)

\(m=98\cdot5a+36.5\cdot3a=5.995\left(g\right)\)

\(\Rightarrow a=0.01\)

\(n_{H_2SO_4}=0.05\left(mol\right),n_{HCl}=0.03\left(mol\right)\)

\(b.\)

\(n_{H_2SO_4}=0.025\left(mol\right),n_{HCl}=0.015\left(mol\right)\)

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

\(n_B=x+y=0.025+0.015=0.04\left(mol\right)\left(1\right)\)

\(m_B=28x+44y=2.16\left(g\right)\left(2\right)\)

\(\left(1\right),\left(2\right):\) Không biết sao tới chổ này số mol âm mất em ơii

a. Đặt \(5a\left(mol\right)=n_{H_2SO_4}\rightarrow3a\left(mol\right)=n_{HCl}\)

\(m_{hh}=m_{H_2SO_4}+m_{HCl}\)

\(\rightarrow5,995=5a.98+3a.36,5\)

\(\rightarrow a=0,01mol\)

\(\rightarrow\hept{\begin{cases}n_{H_2SO_4}=0,05mol\\n_{HCl}=0,03mol\end{cases}}\)

b. \(M_A=\frac{5,995}{0,05+0,03}=74,9375g/mol\)

\(\rightarrow n_A=\frac{2,9975}{74,9375}=0,04mol\)

Vì \(V_A=V_B\rightarrow n_A=n_B=0,04mol\)

Đặt \(\hept{\begin{cases}x=n_{CO}\\y=n_{CO_2}\end{cases}}\)

\(\rightarrow x+y=n_B=0,04\left(1\right)\)

Vì \(m_B=2,16=m_{CO}+m_{CO_2}\)

\(\rightarrow28x+44y=2,16\left(2\right)\)

Từ (1) và (2) => x = -0,025 và y = 0,065

a) Gọi $n_{CO_2} = a(mol) ; n_{SO_2} = b(mol)$
Ta có : 

$a + b = \dfrac{8,96}{22,4} = 0,4(mol)$

$\dfrac{44a + 64b}{a + b} = 27.2$

Suy ra : a = b = 0,2$
$V_{CO_2} = V_{SO_2} = 0,2.22,4 = 4,48(lít)$

b) Theo PTHH : $n_{K_2SO_3} = n_{SO_2} = 0,2(mol)$

$\Rightarrow m_{K_2SO_3} = 0,2.158 = 31,6(gam)$

Gọi $n_{K_2CO_3} = x(mol) ; n_{Na_2CO_3} = y(mol)$

$\Rightarrow 138x + 106y + 31,6 = 56(1)$

$n_{CO_2} = x + y = 0,2(2)$

Từ (1)(2) suy ra : x = y = 0,1

$m_{K_2CO_3} = 0,1.138 = 13,8(gam) ; m_{Na_2CO_3} = 0,1.106 = 10,6(gam)$

a/ nHCl = 5,6 / 22,4 = 0,25 mol

=> m_{hỗn hợp} = 0,3 x 80 + 0,25 x 36,5 = 33.125 gam

b/ n_SO2 = 16 / 64 = 0,25 mol

=> V_{hỗn hợp}(đktc) = ( 0,25 + 0,15 ) x 22,4 = 8,96 lít

\(V_{hh}=\left(0,1+0,2+0,1\right).22,4=8,96\left(l\right)\)