a)
nBa(OH)2 = 0,1.0,1=0,01 mol
nKOH = 0,1.0,1=0,01 mol
nOH- = nBa(OH)2 .2 + nKOH = 0,01.2+0,01 =0,03 mol
P[OH]=0,82
=> pH = 14-p[OH] = 13,18

\(a.m_{HCl}=100.10\%+150.20\%=40\left(g\right)\\ C\%_{ddHCl}=\dfrac{40}{100+150}.100=16\%\\ b.n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{10,95\%.100}{36,5}=0,3\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,3}{1}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,3-2.0,1=0,1\left(mol\right)\\ m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\\ n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\\ m_{MgCl_2}=0,1.95=9,5\left(g\right)\\ m_{ddsau}=2,4+100-0,1.2=102,2\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{3,65}{102,2}.100\approx3,571\%\)

\(C\%_{ddMgCl_2}=\dfrac{9,5}{102,2}.100\approx9,295\%\)

a, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, \(n_{HCl}=2n_{Mg}=0,2\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,2}{2}=0,1\left(l\right)\)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PTHH :

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

0,1        0,2                        0,1 

\(a,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(b,V=\dfrac{n}{C_M}=\dfrac{0,2}{2}=0,1\left(l\right)\)

Mg+2HCl->MgCl₂+H₂

x------------------------x

Zn+2HCl->ZnCl₂+H₂

y-----------------------y

=>Ta có :\(\left\{{}\begin{matrix}24x+65y=15,4\\x+y=0,3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0,1mol\\y=0,2mol\end{matrix}\right.\)

=>m _Mg=0,1.24=2,4g

=>m _Zn=0,2.65=13g

b)V_HCl=\(\dfrac{0,1.2+0,2.2}{2}\)=0,3l=300ml

Đáp án B.

• Cách 1:

gọi x, y lần lượt là số mol của Fe và Mg trong hỗn hợp

số mol H2 là 

theo bài ra ta có hệ phương trình

khối lượng muối khan là :

m = mFeCl2 + mMgCl2 = x(56 + 71) + y(24 + 71)

m = 56x + 24y + 71(x + y)

m = 20 + 71.0,5 = 55,5(g)

• Cách 2:

nmuối = 

Khối lượng muối = m(kim loại) + m(gốc clorua) = 20,0 + 71. 0,5 = 55,5 (g)

Đặt \(n_{Mg}=x(mol);n_{Al}=y(mol)\Rightarrow 24x+27y=7,8(1)\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow x+1,5y=0,4(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ a,m_{Mg}=0,1.24=2,4(g)\\ m_{Al}=7,8-2,4=5,4(g)\\ b,\Sigma n_{HCl}=2x+3y=0,8(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,8}{2}=0,4(l)\)

\(c,PTHH:MgCl_2+2NaOH\to Mg(OH)_2\downarrow+2NaCl\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow+3NaCl\\ n_{MgCl_2}=n_{Mg(OH)_2}=x=0,1(mol)\\ n_{Al(OH)_3}=n_{AlCl_3}=y=0,2(mol)\\ \Rightarrow m_{Mg(OH)_2}=0,1.58=5,8(g)\\ m_{Al(OH)_3}=0,2.78=15,6(g)\\ m_{kết tủa}=5,8+15,6=21,4(g)\)

a.Mg+2HCl->MgCl2+H2

b.nMg=0,1(mol)=>nHCl=0,2(mol)

=>VHCl=0,2/2=0,1(l)=100(ml)

nMgCl2=0,1(mol)

=>mMgCl2=0,1*95=9,5(g)

c.nHCl pư=0,1(mol)

=>nMg dư=0,1-0,05=0,05(mol)

Mg+H2SO4->MgSO4+H2

nH2SO4=0,05(mol)=>mH2SO4=0,05*98=4,9(g)

mdd H2SO4=\(\dfrac{4,9\cdot100}{10}=49\left(g\right)\)