a) \(n_{H_2}=0,2\left(mol\right)\)

Bảo toàn nguyên tố H : \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)

=> \(n_{Cl^-}=0,4\left(mol\right)\)

=> \(m_{muối}=m_{KL}+m_{Cl^-}=20+0,4.35,5=34,2\left(g\right)\)

\(n_{H_2}=\dfrac{20.16}{22.4}=0.9\left(mol\right)\)

\(n_{HCl}=2n_{H_2}=2\cdot0.9=1.8\left(mol\right)\Rightarrow m_{HCl}=1.8\cdot36.5=65.7\left(g\right)\)

Định luật bảo toàn khối lượng : 

\(m_{kl}+m_{HCl}=m_{Muối}+m_{H_2}\)

\(\Rightarrow m_{Muối}=29.4+65.7-1.8=93.3\left(g\right)\)

\(n_{H_2SO_4}=n_{H_2}=a(mol)\\ BTKL:\\ m_{hh}+m_{H_2SO_4}=m_{muối}+m_{H_2}\\ 29+98.a=86,6+2.a\\ \to a=0,6(mol)\\ V_{H_2}=0,6.22,4=13,44(l)\)

Đáp án C

nCO2=0.2 nên nHCl=0.4

=> m muối=20.6+0.4*36.5-0.2*44-0.2*18=22.8

Đáp án : D

Giả sử kim loại M có hóa trị x

=> muối có dạng M₂(SO₄)_x

=> m S O 4 p ư  = m_muối – m_KL = 49,4592g

=>  n S O 4 p ư   = 0,5152 mol

=> M = 20x (g)

Nếu x = 2 => M = 40g => Ca

nSO2= 0,15(mol)

PTHH: Mg + 2 H2SO4(đ) -to> MgSO4 + SO2 + 2 H2O

x________2x______________x_______x(mol)

2 Fe + 6 H2SO4(Đ) -to-> Fe2(SO4)3 + 3 SO2 +6 H2O

y____3y__________0,5y___________1,5y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}24x+56y=5,2\\x+1,5y=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,03\\y=0,08\end{matrix}\right.\)

=> m(muối)= mMgSO4+ mFe2(SO4)3= 120.x+0,5.y.400=120.0,03+0,5.0,08.400=19,6(g

=>CHỌN B

\(n_{XCl_3}=\dfrac{a}{M_X+106,5}\left(mol\right)\)

PTHH: 2X + 6HCl --> 2XCl₃ + 3H₂

=> \(n_X=\dfrac{a}{M_X+106,5}\left(mol\right)\)

\(n_{X_2\left(SO_4\right)_3}=\dfrac{b}{2.M_X+288}\left(mol\right)\)

PTHH: 2X + 3H₂SO₄ --> X₂(SO₄)₃ + 3H₂

=> \(n_X=\dfrac{b}{M_X+144}\left(mol\right)\)