n_Al = 5.4 / 27 = 0.2 (mol)

2Al + 6HCl => 2AlCl₃ + 3H₂

0.2......0.6............0.2.......0.3

a) V_H2 = 0.3 * 22.4 = 6.72 (l) 

b) m_AlCl3 = 0.2 * 133.5 = 26.7 (g) 

c) VddHCl = 0.6 / 1.5 = 0.4 (l) 

d) C_MAlCl3 = 0.2 / 0.4 = 0.5 (M) 

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{HCl}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\end{matrix}\right.\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

a, Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=1,2\left(mol\right)\)

\(\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)

b, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

\(n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,4\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)

a, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

\(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3\%}=200\left(g\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right);n_{HCl}=\dfrac{365.10\%}{36,5}=1\left(mol\right)\\PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{1}{2}>\dfrac{0,1}{1}\Rightarrow HCldư\\ n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)

a, \(KOH+HCl\rightarrow KCl+H_2O\)

b, \(n_{KOH}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(n_{KCl}=n_{KOH}=0,2\left(mol\right)\Rightarrow m_{KCl}=0,2.74,5=14,9\left(g\right)\)

c, \(n_{HCl}=n_{KOH}=0,2\left(mol\right)\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{7,3}{40\%}=18,25\left(g\right)\)

Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

____0,5____________0,5 (mol)

a, \(m_{CuCl_2}=0,5.135=67,5\left(g\right)\)

b, Có: m _{dd sau pư} = m_CuO + m _{dd HCl} = 40 + 200 = 240 (g)

\(\Rightarrow C\%_{CuCl_2}=\dfrac{67,5}{240}.100\%=28,125\%\)

Bạn tham khảo nhé!

nAl = 5.4/27 = 0.2 mol 

2Al + 3H2SO4 => Al2(SO4)3 + 3H2 

0.2______________0.1________0.3 

VH2 = 0.3*22.4 = 6.72 (l) 

mAl2(SO4)3 = 0.1*342 = 34.2 g 

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{25,55}{36,5}=0,7\left(mol\right)\\a. 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.Vì:\dfrac{0,2}{2}< \dfrac{0,7}{6}\\ \Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,7-\dfrac{6}{2}.0,2=0,1\left(mol\right)\\ n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ m_{H_2}=0,3.2=0,6\left(g\right)\\ m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\\ m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\)

$Fe+2HCl\rightarrow FeCl_2+H_2$

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

a) Khối lượng dung dịch HCl tối thiểu cần dùng:

Theo phương trình:

$n_{HCl}=2n_{Fe}=0,4\ mol$

$m_{HCl}=0,4\cdot36,5=14,6\ gam$

$m_{dd\ HCl}=\dfrac{14,6\cdot100}{7,3}=200\ gam$

b) Khối lượng muối tạo thành:

$n_{FeCl_2}=n_{Fe}=0,2\ mol$

$m_{FeCl_2}=0,2\cdot127=25,4\ gam$

c) Nồng độ phần trăm dung dịch muối khi axit vừa đủ:

$n_{H_2}=n_{Fe}=0,2\ mol$

$m_{H_2}=0,2\cdot2=0,4\ gam$

$m_{dd\ sau\ phản\ ứng}=200+11,2-0,4=210,8\ gam$

$C\%_{FeCl_2}=\dfrac{25,4}{210,8}\cdot100\%\approx12,05\%$

d) Cho $11,2$ gam Fe vào $300$ gam dung dịch HCl $7,3\%$:

$m_{HCl}=\dfrac{300\cdot7,3}{100}=21,9\ gam$

$n_{HCl}=\dfrac{21,9}{36,5}=0,6\ mol$

Theo phương trình, $n_{HCl\ phản\ ứng}=0,4\ mol$.

Vì $0,6>0,4$ nên HCl dư.

$n_{HCl\ dư}=0,6-0,4=0,2\ mol$

$m_{HCl\ dư}=0,2\cdot36,5=7,3\ gam$

$m_{FeCl_2}=25,4\ gam$

$m_{dd\ sau\ phản\ ứng}=300+11,2-0,4=310,8\ gam$

Dung dịch sau phản ứng gồm $FeCl_2$ và $HCl$ dư.

$C\%_{FeCl_2}=\dfrac{25,4}{310,8}\cdot100\%\approx8,17\%$

$C\%_{HCl}=\dfrac{7,3}{310,8}\cdot100\%\approx2,35\%$

Vậy:

$a)\ m_{dd\ HCl}=200\ gam$

$b)\ m_{FeCl_2}=25,4\ gam$

$c)\ C\%_{FeCl_2}\approx12,05\%$

$d)\ C\%_{FeCl_2}\approx8,17\%;\quad C\%_{HCl}\approx2,35\%$