16g

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Đáp án A

m_muối = m_KL + m_{gốc axit} => 19,2 = 12,1 + m_Cl-

 ₌ n_HCl

_.100%

=> m = 73 g

nFe=5,6/56=0,1(mol)

Fe+2HCl--->FeCl2+H2

0,1__0,2_____0,1___0,1

mH2=0,1.2=0,2(g)

mdung dịch=5,6+200-0,2=205,4(g)

=>mHCl=>C%HCl

mFeCl2=0,1.127=12,7(g)

=>C%FeCl2=12,7/205,4.100%~6,183%

3/

nFe= 8.4/56=0.15mol

Fe +2HCl --> FeCl2 + H2

0.15__0.3____________0.15

VH2= 0.15*22.4=3.36l

mHCl= 0.3*36.5=10.95g

mddHCl= 10.96*100/10.95=100g

4/ nZn= 3.25/65=0.05 mol

Zn + 2HCl --> ZnCl2 + H2

0.05_______________0.05

nCuO= 12/80=0.15 mol

CuO + H2 -to-> Cu + H2O

Bđ: 0.15___0.05

Pư: 0.05___0.05___0.05

Kt: 0.1_____0_____0.05

mCu= 0.05*64=3.2g

a)

Gọi $n_{Mg} = a ; n_{Al} = b \Rightarrow 24a + 27b = 5,1(1)$
$Mg + 2HCl \to MgCl_2  + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$

Ta có :

$n_{H_2} = a + 1,5b = \dfrac{5,6}{22,4} = 0,25(2)$
Từ (1)(2) suy ra a = b = 0,1

$\%m_{Mg} = \dfrac{0,1.24}{5,1}.100\%  =47,06\%$

$\%m_{Al} = 52,94\%$

b)

$n_{HCl} = 2n_{H_2} = 0,5(mol)$
$m_{dd\ HCl} = \dfrac{0,5.36,5}{10\%} = 182,5(gam)$

c)

$MgCl_2 + 2NaOH \to Mg(OH)_2 + 2NaCl$
$AlCl_3 + 3NaOH \to Al(OH)_3 + 3NaCl$

$Al(OH)_3 + NaOH \to NaAlO_2 + 2H_2O$

$n_{Mg(OH)_2} = a = 0,1(mol)$
$\Rightarrow m_{kết\ tủa}  = 0,1.58 = 5,8(gam)$

Ta có:

\(Mg+2HCl\rightarrow MgCl_2+H_2\) ; \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Đặt số mol Mg và Al lần lượt là a và b (a,b>0)
theo bài ra ta có hệ 

\(\left\{{}\begin{matrix}24a+27b=5,1\\a+1,5b=\dfrac{5,6}{22,4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\%Mg=\dfrac{0,1\times24}{5,1}=47,06\%\Rightarrow\%Al=100\%-47,06\%=52,94\%\)

Theo PT có \(n_{HCl}=2n_{Mg}+3n_{Al}=2\times0,1+3\times0,1=0,5\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,5\times36,5=18,25\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{18,25}{10\%}=182,5\left(g\right)\)

\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)

\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)

+ Với NaOH vừa đủ

\(a=m_{Mg\left(OH\right)_2}+m_{Al\left(OH\right)_3}=0,1\times58+0,1\times78=13,6\left(g\right)\)

+ Với NaOH dư có thêm PT

\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)

\(\Rightarrow a=m_{Mg\left(OH\right)_2}=0,1\times58=5,8\left(g\right)\)

- Tính n H2 = cách bảo toàn e =» n hcl pư =» m dd hcl pư

- Bảo toàn nguyên tố Fe để tím n FeCl2 =» m FeCl2 (dd B)

C% dd B = m FeCl 2 / (m Fe + m dd HCl)

\(n_{H2}=\frac{1}{2}n_{HCl}=0,25\left(mol\right)=>m_{H2}=0,5\left(g\right)\)

m dd sau pư =\(m_{KL}+m_{ddHCl}-m_{Crắn}-m_{H2}\)

\(=10,8+91,25-0,5-12,8=88,75\left(g\right)\)

\(C\%_{MgCl2}=\frac{0,1.95}{88,75}.100\%=10,7\%\)

\(C\%_{AlCl3}=\frac{0,15.133,5}{88,75}.100\%=22,56\%\)

\(\text{Câu 1: a) }pthh:R+Cl_2\overset{t^0}{\rightarrow}RCl_2\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }R\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }R+71\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }6,5\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }13,6\\ \Rightarrow13,6R=6,5\left(R+71\right)\\ \Rightarrow13,6R=6,5R+461,5\\ \Rightarrow7,1R=77,5\\ \Rightarrow7,1R=461,5\\ \Rightarrow R=65\left(Zn\right)\)

Vậy kim loại cần tìm là Kẽm \(\left(Zn\right)\)

b) \(pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\left(2\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{6.5}{65}=0,1\left(mol\right)\)

Theo \(pthh\left(2\right):n_{HCl}=2n_{Zn}=2\cdot0,1=0,2\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)