Ta có: nMgCl2=\(\dfrac{38}{95}\)=0,4(mol);

nCO2=\(\dfrac{6,72}{22,4}\)=0,3(mol)

          MgCO3 + 2HCl → MgCl2 + CO2 + H2O

(mol)    0,3 ←                      0,3  ← 0,3

           MgO + 2HCl → MgCl2 + H2O

(mol)     0,1 ←                  0,1

\(\left\{{}\begin{matrix}mMgO=0,1.40=4g\\mMgCO3=0,3.84=25,2\end{matrix}\right.\)

=>%mMgO=\(\dfrac{4}{29,2}\).100=13,7%

=>%m MgCO4=86,3%

a/ \(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(MgCO_3+H_2SO_4\rightarrow MgSO_4+H_2O+CO_2\)

\(0,5---0,5----0,5---0,5-0,5\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(b---b----b-----b\)

\(\Rightarrow m_{MgCO_3}=0,5.\left(24+12+16.3\right)=42\left(g\right)\)

\(\dfrac{m_{MgCO_3}}{m_{MgO}}=\dfrac{7}{3}\Rightarrow m_{MgO}=42.\dfrac{3}{7}=18\left(g\right)\Rightarrow n_{MgO}=b=0,45\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}=0,45+0,5=0,95\left(mol\right)\) \(\Rightarrow m_{dd}=\dfrac{0,95.98}{0,05}=1862\left(g\right)\)

a) 

\(n_{MgCl_2}=\dfrac{38}{95}=0,4\left(mol\right)\)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: MgCO₃ + 2HCl --> MgCl₂ + CO₂ + H₂O

                0,3<------0,6<------0,3<----0,3

            MgO + 2HCl --> MgCl₂ + H₂O

               0,1<---0,2<------0,1

=> \(\left\{{}\begin{matrix}m_{MgO}=0,1.40=4\left(g\right)\\m_{MgCO_3}=0,3.84=25,2\left(g\right)\end{matrix}\right.\)

b) \(m_{HCl}=\left(0,6+0,2\right).36,5=29,2\left(g\right)\)

=> \(m_{dd.HCl}=\dfrac{29,2.100}{20}=146\left(g\right)\)

\(a.PTHH:\)

\(Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)

\(MgO+2HCl--->MgCl_2+H_2O\left(2\right)\)

b. ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgO}=12-0,2.24=7,2\left(g\right)\)

\(\Rightarrow\%_{MgO}=\dfrac{7,2}{12}.100\%=60\%\)

c. Ta có: \(n_{hh}=0,2+\dfrac{7,2}{40}=0,38\left(mol\right)\)

Theo PT_(1,2): \(n_{HCl}=2.n_{hh}=2.0,38=0,76\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,76.36,5=27,74\left(g\right)\)

\(\Rightarrow m_{dd_{HCl}}=138,7\left(g\right)\)

\(\Rightarrow V_{dd_{HCl}}=126\left(ml\right)\)

Gọi \(\left\{{}\begin{matrix}n_{H_2}=a\left(mol\right)\\n_{CO_2}=b\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a+b=\dfrac{4,704}{22,4}=0,21\\\overline{M}=\dfrac{2a+44b}{a+b}=12,5.2=25\end{matrix}\right.\)

=> a = 0,095 (mol); b = 0,115 (mol)

Đặt n_HCl = x (mol)

\(n_{MgCl_2}=\dfrac{12,825}{95}=0,135\left(mol\right)\)

Bảo toàn Cl: \(n_{CaCl_2}=\dfrac{x-0,27}{2}\left(mol\right)\)

Bảo toàn H: \(n_{H_2O}=\dfrac{x-2.0,095}{2}=\dfrac{x-0,19}{2}\left(mol\right)\)

BTKL: 

\(m_{hh\left(bđ\right)}+m_{HCl}=m_{MgCl_2}+m_{CaCl_2}+m_{H_2}+m_{CO_2}+m_{H_2O}\)

=> \(19,02+36,5x=12,825+\dfrac{x-0,27}{2}.111+0,095.2+0,115.44+\dfrac{x-0,19}{2}.18\)

=> x = 0,63 (mol)

=> \(n_{CaCl_2}=0,18\left(mol\right)\)

=> m_CaCl2 = 0,18.111 = 19,98 (g)

Đáp án A

n khí = 0,42 mol

nMgCl₂ = 0,27 mol

Quy đổi hỗn hợp đầu về: Ca (x mol), Mg (0,27 mol), C (0,23 mol), O (y mol)

m _{hỗn hợp} = 40x+0,27.24+0,23.12+16y = 38,04 (1)

BT e: 2n_Ca + 2n_Mg + 4n_C = 2n_O + 2n_H2 => 2x + 0,27.2 + 0,23.4 = 2y + 0,19.2 (2)

Giải (1) và (2) => x = 0,36; y = 0,9

=> mCaCl₂ = 0,36.111 = 39,96 gam

Đáp án là D. 19,98

Giải thích: 

n khí = 0,42 mol

nMgCl₂ = 0,27 mol

CO₂:    44                    23                    0,23

                        25                    =

H₂:       2                      19                    0,19

Quy đổi hỗn hợp đầu về: Ca (x mol), Mg (0,27 mol), C (0,23 mol), O (y mol)

m _{hỗn hợp} = 40x+0,27.24+0,23.12+16y = 38,04 (1)

BT e: 2n_Ca + 2n_Mg + 4n_C = 2n_O + 2n_H2 => 2x + 0,27.2 + 0,23.4 = 2y + 0,19.2 (2)

Giải (1) và (2) => x = 0,36; y = 0,9

=> mCaCl₂ = 0,36.111 = 39,96 gam

Đáp án A