Chọn A

a) $Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH :

$n_{H_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$

$V_{H_2} = 0,2.22,4 = 4,48(lít)$

b) $n_{HCl} = 2n_{Fe} = 0,4(mol)$

$m_{HCl} = 0,4.36,5 = 14,6(gam)$

c) $2H_2 + O_2 \xrightarrow{t^o}2H_2O$

Theo PTHH : 

$V_{O_2} = \dfrac{1}{2}V_{H_2} = 2,24(lít)$

$n_{H_2O} = n_{H_2} = 0,2(mol)$
$m_{H_2O} = 0,2.18 = 9(gam)$

Bài 1

\(a.n_{H_2}=\dfrac{3,7185}{24,79}=0,15mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,1          0,3            0,1          0,15

\(\%m_{Al}=\dfrac{0,1.27}{10,7}\cdot100=25,23\%\\ \%m_{MgO}=100-25,23=74,77\%\)

\(b.n_{MgO}=\dfrac{10,7-0,1.27}{40}=0,2mol\\ MgO+2HCl\rightarrow MgCl_2+H_2O\)

0,2             0,4

\(V_{ddHCl}=\dfrac{0,3+0,4}{0,5}=1,4l\)

Bài 2:

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 6,5 + 100 - 0,1.2 = 106,3 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)

Bài 2 : 

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{14,6\%.100}{100\%.36,5}=0,4\left(mol\right)\)

a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,1        0,4         0,1        0,1

b) Xét tỉ lệ : \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\Rightarrow HCldư\)

\(\Rightarrow V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

c) \(m_{ddspu}=6,5+100-0,1.2=106,3\left(g\right)\)

\(C\%_{ddHCl\left(dư\right)}=\dfrac{\left(0,4-0,1.2\right).36,5}{106,3}.100\%=6,87\%\)

\(C\%_{ZnCl2}=\dfrac{0,1.136}{106,3}.100\%=12,79\%\)

Đáp án A

Đáp án C

+     n H C l = 0 , 05     m o l ⇒ m m u ố i = m a min + m H C l = 15 + 0 , 05 . 36 , 5 = 16 , 825     g a m

\(n_{Zn}=\dfrac{16}{65}\approx 0,25(mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{Zn}=0,5(mol)\\ \Rightarrow m_{HCl}=0,5.36,5=18,25(g)\)

phản ứng với axit sunfuric nhưng phương trình là HCl :) ?

\(Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{16}{65}=0,25\left(mol\right)\\ n_{HCl}=2n_{Zn}=0,5\left(mol\right)\\ \Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(Fe+2HCl->FeCl_2+H_2\)

0,2----0,4----------0,2--------0,2 ( mol )

n_Fe= 11,2/56=0,2 mol => n_H2 = 0,2 mol

\(2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\)

n_Al = (0,2:3).2=2/15 mol

m_Al= (2/15 ).27=3,6 gam

n_H2SO4 =0,2 mol => n_H2SO4 =0,2.98=19,6 gam

=> mdd_H2SO4 = (19,6/19,6).100=100 gam dung dịch

Vậy...

Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl -> FeCl₂ + H₂ (1)

2Al + 3H₂SO₄ -> Al₂(SO₄)₃ + 3H₂ (2)

Theo đề bài, ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ =>n_{H_2\left(1\right)}=n_{Fe}=0,2\left(mol\right)\\ =>n_{H_2\left(2\right)}=n_{H_2\left(1\right)}=0,2\left(mol\right)\\ =>n_{Al}=\dfrac{2.0,2}{3}=\dfrac{2}{15}\left(mol\right)\\ =>m_{Al}=\dfrac{2}{15}.27=3,6\left(g\right)\\ n_{H_2SO_4}=n_{H_2\left(2\right)}=0,2\left(mol\right)\\ =>m_{H_2SO_4}=0,2.98=19,6\left(g\right)\\ =>m_{ddH_2SO_4}=\dfrac{19,6.100}{19,6}=100\left(g\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂ (1)

2Al + 3H₂SO₄ -> Al₂(SO₄)₃ + 3H₂ (2)

Ta có : n_Fe = \(\dfrac{11,2}{56}\) = 0,2 mol

Cứ 1 mol Fe --> 1 mol H₂ (1)

0,2 mol --> 0,2 mol => n_H2 (2) = 0,2 mol

Ta lại có: Cứ 2 mol Al --> 3 mol H₂SO₄ --> 3 mol H₂

\(\dfrac{2}{15}\) mol <-- 0,2 mol <-- 0,2 mol

=> m_Al = \(\dfrac{2}{15}\times27\) = 3,6 (g)

=> \(m_{H_2SO_4\left(ct\right)}\) = 0,2 x 98 = 19,6 (g)

=> \(m_{dd.Axit}\) = \(\dfrac{19,6}{19,6\%}\) = 100 (g)

Vậy ..............................

1)

a, \(n_{Al}=\dfrac{15,3}{102}=0,15\left(mol\right)\)

PTHH: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

Mol:     0,15         0,9        0,3      

\(m_{ddHCl}=\dfrac{0,9.36,5.100}{20}=164,25\left(g\right)\)

b, m_{dd sau pứ} = 15,3 + 164,25 = 179,55 (g)

c, \(C\%_{ddAlCl_3}=\dfrac{0,3.133,5.100\%}{179,55}=22,31\%\)

2)

a, \(m_{HCl}=54,75.20\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: Al₂O₃  + 6HCl → 2AlCl₃ + 3H₂O

Mol:      0,05        0,3          0,1

\(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)

b, m_{dd sau pứ} = 5,1 + 54,75 = 59,85 (g) 

\(C\%_{ddAlCl_3}=\dfrac{0,1.133,5.100\%}{59,85}=22,31\%\)