a, \(m_{HCl}=150.14,6\%=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{Zn}=n_{ZnCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)

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

c, Ta có: m _{dd sau pư} = 19,5 + 150 - 0,3.2 = 168,9 (g)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,3.136}{168,9}.100\%\approx24,16\%\)

a) 

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

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2-->0,4----->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) m_HCl = 0,4.36,5 = 14,6 (g)

=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)

c)

m_{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

m_ZnCl2 = 0,2.136 = 27,2 (g)

=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, \(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

d, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)

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

⇒ m _{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{27,2}{212,6}.100\%\approx12,79\%\)

Bài 1: Ta có: \(n_{H_2}=\dfrac{1,008}{22,4}=0,045\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

__0,045__0,09____0,045___0,045 (mol)

a, Ta có: \(a=m_{Mg}=0,045.24=1,08\left(g\right)\)

b, \(V_{ddHCl}=\dfrac{0,09}{0,1}=0,9\left(l\right)\)

c, \(C_{M_{MgCl_2}}=\dfrac{0,045}{0,9}=0,05M\)

Bài 2:

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

a, Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

⇒ 24x + 56y = 5,2 (1)

Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=x+y\left(mol\right)\)

⇒ x + y = 0,15 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{5,2}.100\%\approx46,2\%\\\%m_{Fe}\approx53,8\%\end{matrix}\right.\)

b, Ta có: \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,3}{1}=0,3\left(l\right)\)

Bạn tham khảo nhé!

a)

$n_{Al} = 0,3(mol)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

Theo PTHH : 

$n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,45.98}{12,25\%} = 360(gam)$

b)

$n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$

c)

$n_{Al_2(SO_4)_3} = 0,15(mol)$
$m_{dd\ sau\ pư} = 8,1 + 360 - 0,45.2 = 367,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{367,2}.100\% = 14\%$

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

pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

        1         2             1           1

       0,2     0,4           0,2         0,2

a) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(m_{ddHCl}=\dfrac{14,6.100}{14,6}=100\left(g\right)\)

\(n_{H2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

b) \(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{FeCl2}=0,2.127=25,4\left(g\right)\)

\(m_{ddspu}=11,2+100-\left(0,2.2\right)=110,8\left(g\right)\)

\(C_{FeCl2}=\dfrac{25,4.100}{110,8}=22,92\)⁰/₀

 Chúc bạn học tốt

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, Có lẽ phần này đề hỏi khối lượng sắt chứ bạn nhỉ?

 \(n_{ZnCl_2}=0,4.2=0,8\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{FeCl_2}=0,8\left(mol\right)\Rightarrow m_{Fe}=0,8.56=44,8\left(g\right)\)

c, \(n_{H_2}=n_{FeCl_2}=0,8\left(mol\right)\Rightarrow V_{H_2}=0,8.22,4=17,92\left(l\right)\)

d, \(n_{HCl}=2n_{FeCl_2}=1,6\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,4}=4\left(M\right)\)

a) $2Al + 6HCl \to 2AlCl_3 + 3H_2$

b) n Al = 8,1/27 = 0,3(mol)

Theo PTHH : 

n H2 = 3/2 n Al = 0,45(mol)

V H2 = 0,45.22,4 = 10,08(lít)

c) n AlCl3 = n Al = 0,3(mol)

m AlCl3 = 0,3.133,5 = 40,05(gam)

d) n HCl = 3n Al = 0,9(mol)

m dd HCl = 0,9.36,5/7,3% = 450(gam)

Sau phản ứng : 

m dd = 8,1 + 450  -0,45.2 = 457,2(gam)

C% AlCl3 = 40,05/457,2  .100% = 8,76%