MgCO3+2HCl->MgCl2+H2O+CO2

0,25--------0,5

n MgCO3=\(\dfrac{21}{84}\)=0,25 mol

=>VHCl=\(\dfrac{0,5}{2}\)=0,25 l=250ml

=>B

B. 0,25 lít

\(n_{MgCO_3}=\dfrac{21}{84}=0,25mol\)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

 0,25          0,5                                             ( mol )

\(V_{HCl}=\dfrac{0,5}{2}=0,25l\)

=> Chọn B

\(n_{MgCO_3}=\dfrac{1,68}{84}=0,02\left(mol\right)\)

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

______0,02--->0,04

=> \(V_{ddHCl}=\dfrac{0,04}{2}=0,02\left(l\right)\)

Số mol oxi là 8,64/16=0,54 (mol), suy ra số mol CO₃^2- trong X là 0,54/3=0,18 (mol) bằng số mol CO₂ và H₂O sinh ra sau phản ứng (2H⁺ + CO₃^2- \(\rightarrow\) CO₂ + H₂O), suy ra số mol HCl phản ứng là 2n_nước=0,18.2=0,36 (mol).

BTKL: m=28,12+0,36.36,5-0,18.44-0,18.18=30,1 (g).

\(n_{HCl}=\dfrac{100.14,6\%}{36,5}=0,4\left(mol\right)\\ n_{MgCO_3}=\dfrac{50}{84}=\dfrac{25}{42}\left(mol\right)\\ PTHH:MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\\ Vì:0,4:2< \dfrac{25}{42}:1\\ \Rightarrow MgCO_3dư\\ \Rightarrow ddsau:MgCl_2\\n_{MgCO_3\left(p.ứ\right)}=n_{CO_2}= n_{MgCl_2}=\dfrac{n_{HCl}}{2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ m_{ddsau}=m_{MgCO_3\left(p.ứ\right)}+m_{ddHCl}-m_{CO_2}=0,2.84+100-0,2.44=108\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{108}.100\approx17,593\%\%\)

28.

\(n_{MgCO_3}=\dfrac{4,2}{84}=0,05(mol)\\ MgCO_3+2HCl\to MgCl_2+H_2O+CO_2\uparrow\\ \Rightarrow n_{HCl}=0,1(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,1}{2}=0,05(l)\)

29.

\(Zn+H_2SO_4\to ZnSO_4+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ \Rightarrow \%_{Zn}=\dfrac{0,1.65}{10,5}.100\%=62\%\\ \Rightarrow \%_{Cu}=100\%-62\%=38\%\)

30.

Đặt KL là R

\(n_{HCl}=2.0,1=0,2(mol)\\ 2R+O_2\xrightarrow{t^o}2RO\\ RO+2HCl\to RCl_2+H_2O\\ \Rightarrow n_RO=0,1(mol)\\ \Rightarrow M_RO=\dfrac{4}{0,1}=40(g/mol)\\ \Rightarrow M_R=40-16=24(g/mol)(Mg)\)

a

\(n_{MgCO_3}=\dfrac{12,6}{84}=0,15\left(mol\right)\)

\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)

0,15                                                     0,15

\(V_{CO_2}=0,15.22,4=3,36\left(l\right)\)

--> A

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