Đáp án C

Đáp án A

Gọi n là hóa trị của kim loại A

Coi $m_{dd\ H_2SO_4} = 98(gam)$

$\Rightarrow n_{H_2SO_4} = \dfrac{98.10\%}{98} = 0,1(mol)$

$2A + nH_2SO_4 \to A_2(SO_4)_n + nH_2$

Theo PTHH :

$n_{H_2} = n_{H_2SO_4} = 0,1(mol)$

$n_A = \dfrac{2}{n}.n_{H_2SO_4} = \dfrac{0,2}{n}(mol)$

Suy ra : 

$m_{dd} = m_A + m_{dd\ H_2SO_4} - m_{H_2} = \dfrac{0,2A}{n} + 97,8(gam)$

$m_{muối} = \dfrac{0,1}{n}.(2A + 96) = \dfrac{0,2A}{n} + 9,6(gam)$

Suy ra : 

\(C\% = \dfrac{\dfrac{0,2A}{n} +9,6}{\dfrac{0,2A}{n} + 97,8}.100\% = 14,7\%\)

=> 0,2A/n =5,6

=> A = 28n

Với n = 2 thì A = 56(Fe)

\(X+H_2SO_4\rightarrow XSO_4+H_2\\Tacó: n_X=n_{XSO_4}\\ \Rightarrow\dfrac{7,2}{X}=\dfrac{36}{X+96}\\ \Rightarrow X=24\left(Magie-Mg\right)\)

\(n_X=\dfrac{7,2}{M_X}\left(mol\right)\)

PTHH: X + H₂SO₄ --> XSO₄ + H₂

____\(\dfrac{7,2}{M_X}\)----------->\(\dfrac{7,2}{M_X}\)

=> \(\dfrac{7,2}{M_X}\left(M_X+96\right)=36=>M_X=24\left(g/mol\right)\)

=> X là Mg (Magie)

Gọi x là số mol của kim loại X

PTHH: X + H₂SO₄  XSO₄ + H₂

Theo định luật bảo toàn khối lượng ta có:

 mX + mH₂SO₄ = _m muối + ^mH₂

 7,2  + 98x         = 36        + 2x

                 x = 0,3 (mol)

M_X = 7,2 : 0,3 = 24 (g/mol)                    

   X là magie (Mg)

\(n_{H_2SO_4}=\dfrac{78,4.20}{100.98}=0,16\left(mol\right)\)

PTHH: X + H₂SO₄ --> XSO₄ + H₂

____0,16<--0,16--->0,16-->0,16

=> 0,16.M_X + 78,4 - 0,16.2 = 88,48

=> M_X = 65 (g/mol)

=> X là Zn

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

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

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

Câu 2:

\(n_{MgBr_2}=\dfrac{14,72}{184}=0,08\left(mol\right)\\ Mg+Br_2\rightarrow MgBr_2\\ n_{Mg}=n_{Br_2}=n_{MgBr_2}=0,08\left(mol\right)\\ a=m_{Mg}=24.0,08=1,92\left(g\right)\\ m_{Br_2}=160.0,08=12,8\left(g\right)\)

Câu 1:

\(n_{AlBr_3}=\dfrac{106,8}{267}=0,4\left(mol\right)\\ 2Al+3Br_2\rightarrow2AlBr_3\\ n_{Al}=n_{AlBr_3}=0,4\left(mol\right)\\ n_{Br_2}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\\ a=m_{Al}=0,4.27=10,8\left(g\right)\\ m_{Br_2}=160.0,6=96\left(g\right)\)

2Al+3Br₂-to>2AlBr₃

0,4-----0,6-------0,4 mol

n AlBr3=\(\dfrac{106,8}{267}\)=0,4 mol

=>a=mAl=0,4.27=10,8g

=>mBr₂=0,6.160=96g

Câu 2

Mg+Br₂->MgBr₂

0,08--0,08---0,08 mol

n _MgBr2=\(\dfrac{14,72}{184}\)=0,08 mol

=>a=mMg=0,08.24=1,92g

=>m _Br2=0,08.160=12,8g

\(PTHH:A+2HCl\to ACl_2+H_2\\ \Rightarrow n_{A}=n_{ACl_2}\\ \Rightarrow \dfrac{27,4}{M_A}=\dfrac{41,6}{M_A+71}\\ \Rightarrow 41,6M_A=27,4M_A+1945,4\\ \Rightarrow 14,2M_A=1945,4\\ \Rightarrow M_A=137(g/mol)\)

Vậy A là Bari (Ba)

Gọi CT oxit : \(R_2O_x\)

Ta có \(n_{R_2O_x}=\dfrac{3,2}{2R+16x}\left(mol\right)\)

PTHH: \(R_2O_x+xH_2SO_4\rightarrow R_2\left(SO_4\right)_x+xH_2O\) (1) 

\(\dfrac{3,2}{2R+16x}\)---->\(\dfrac{3,2}{2R+16x}.x\)-->\(\dfrac{3,2}{2R+16x}\) (mol)

\(H_2SO_{4\left(dư\right)}+CaCO_3\rightarrow CaSO_4+CO_2+H_2O\) (2)

\(n_{CO_2}=\dfrac{0,244}{22,4}=0,01\left(mol\right)\)

Theo PT (2) : \(n_{CO_2}=n_{CaSO_4}=n_{H_2SO_4dư}=0,01\left(mol\right)\)

Ta có : \(m_{muối}=\text{​​}\dfrac{3,2}{2R+16x}.\left(2R+96x\right)+0,01.136=9,36\left(g\right)\)

\(\dfrac{3,2}{2R+16x}\).(2R+96x)=8 

Lập bảng : 

=> Oxit là Fe2O3

\(\Sigma n_{H_2SO_4}=0,02.3+0,01=0,07\left(mol\right)\)

=> \(C\%_{H_2SO_4}=\dfrac{0,07.98}{200}.100=3,43\%\)