\(n_{ZnO}=\dfrac{1,62}{81}=0,02\left(mol\right)\)

\(m_{HCl}=109,5.10\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: ZnO + 2HCl → ZnCl₂ + H₂O

Mol:     0,02      0,3        0,02

Ta có: \(\dfrac{0,02}{1}< \dfrac{0,3}{1}\) ⇒ ZnO hết, HCl dư

m_{dd sau pứ} = 1,62+109,5 = 111,12 (g)

\(C\%_{ddZnCl_2}=\dfrac{0,02.136.100\%}{111,12}=2,45\%\)

\(C\%_{ddHCl}=\dfrac{\left(0,3-0,04\right).36,5.100\%}{111,12}=8,54\%\)

\(n_{ZnO}=\dfrac{1,62}{81}=0,02\left(mol\right)\)

\(m_{HCl}=109,5.10\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: ZnO + 2HCl → ZnCl₂ + H₂

Mol:     0,02    0,04        0,02      0,02

Ta có: \(\dfrac{0,02}{1}< \dfrac{0,3}{2}\) ⇒ ZnO hết, HCl dư

m_{dd sau pứ} = 1,62+109,5-0,02.2 = 111,08  (g)

\(C\%_{ZnCl_2}=\dfrac{0,02.136.100\%}{111,08}=2,45\%\)

\(C\%_{HCldư}=\dfrac{\left(0,3-0,04\right).36,5.100\%}{111,08}=8,54\%\)

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

PTHH :

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

0,3     0,6            0,3       0,3 

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

\(b,m_{HCl}=0,6.36,5=21,9\left(g\right)\)

\(m_{ddHCl}=\dfrac{21,9.100}{21,90}=100\left(g\right)\)

\(c,m_{FeCl_2}=127.0,3=38,1\left(g\right)\)

\(m_{ddFecl_2}=\left(16,8+100\right)-0,3.2=116,2\left(g\right)\)

\(C\%_{ddFeCl_2}=\dfrac{38,1}{116,2}.100\%\approx37,79\%\)

\(n_{Zn}=\dfrac{130}{65}=2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=2mol\\ V_{H_2}=2.22,4=44,8l\\ 1000ml=1l\\ n_{HCl}=2.2=4mol\\ C_{M_{HCl}}=\dfrac{4}{1}=4M\)

\(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{130}{65}=2mol\) 

PTHH: Zn + 2HCl \(\rightarrow\) ZnCl₂ + H₂

TL;       1        2          1          1

mol:      2 \(\rightarrow\) 4

\(m_{HCl}=n.M=4.36,5=146g\) 

đổi 1000 ml= 1l

\(C\%_{ddHCl}=\dfrac{m_{HCl}}{V_{HCl}}.100\%=\dfrac{146}{1}.100=14600\%\)

số hơi lớn em xem lại đề nhé

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(m_{HCl}=245.16,7\%=40,915\left(g\right)\Rightarrow n_{HCl}=\dfrac{40,915}{36,5}=1,121\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{1,121}{2}\), ta được HCl dư.

Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)

⇒ n_{HCl dư} = 1,121 - 0,2 = 0,921 (mol)

Ta có: m _{dd sau pư} = 5,6 +245 - 0,1.2 = 250,4 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,1.127}{250,4}.100\%\approx5,07\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,921.36,5}{250,4}.100\%\approx13,43\%\end{matrix}\right.\)

250ml = 0,25 lít

\(n_{Na_2O}=\dfrac{3,1}{62}=0,05mol\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,05                          0,1      ( mol )

\(C_{MNaOH}=\dfrac{0,1}{0,25}=0,4M\)