\(n_{FeO}=\dfrac{14,4}{72}=0,2\left(mol\right)\)

\(m_{ct}=\dfrac{10,95.200}{100}=21,9\left(g\right)\)

\(n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Pt : \(FeO+2HCl\rightarrow FeCl_2+H_2O|\)

         1           2             1            1

        0,2       0,6            0,2

a) Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,6}{2}\)

             ⇒ FeO phản ứng hết , HCl dư

             ⇒ Tính toán dựa vào số mol của FeO

\(n_{HCl\left(dư\right)}=0,6-\left(0,2.2\right)=0,2\left(mol\right)\)

⇒ \(m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)

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

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

\(m_{ddspu}=14,4+200=214,4\left(g\right)\)

\(C_{FeCl2}=\dfrac{25,4.100}{214,4}=11,85\)⁰/₀

\(C_{HCl\left(dư\right)}=\dfrac{7,3.100}{214,4}=3,4\)⁰/₀

 Chúc bạn học tốt

a,\(n_{FeO}=\dfrac{14,4}{72}=0,2\left(mol\right)\)

 \(m_{HCl}=10,95\%.200=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

PTHH: FeO + 2HCl → FeCl₂ + H₂O

Mol:      0,2      0,4         0,2

Ta có: \(\dfrac{0,2}{1}< \dfrac{0,6}{2}\) ⇒ FeO pứ hết ,HCl dư

m_{HCl dư} = (0,6-0,4).36,5 = 7,3 (g)

b, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

c, m_{dd sau pứ} = 14,4+200 = 214,4 (g)

\(C\%_{ddFeCl_2}=\dfrac{25,4.100\%}{214,4}=11,847\%\)

\(C\%_{ddHCldư}=\dfrac{7,3.100\%}{214,4}=3,4\%\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

a) Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\) \(\Rightarrow n_{HCl}=0,2mol\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2\cdot36,5}{10,95\%}\approx66,67\left(g\right)\)

b) Theo PTHH: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1mol\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Zn}+m_{ddHCl}-m_{H_2}=72,97\left(g\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{13,6}{72,97}\cdot100\%\approx18,64\%\)

a) \(\%m_{MgO}=56,66\%\)

   \(\%m_{Al_2O_3}=43,34\%\)

Ta có: \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

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

______0,2_____0,4_____0,2 (mol)

a, \(m_{CuCl_2}=0,2.135=27\left(g\right)\)

b, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow C\%_{HCl}=\dfrac{14,6}{300}.100\%\approx4,867\%\)

c, Ta có: m _{dd sau pư} = 16 + 300 = 316 (g)

\(\Rightarrow C\%_{CuCl_2}=\dfrac{27}{316}.100\%\approx8,54\%\)

PTHH: \(K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\uparrow\)

a+b+c) Ta có: \(n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{KCl}=0,5\left(mol\right)=n_{HCl}\\n_{K_2SO_3}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_3}=0,25\cdot158=39,5\left(g\right)=a\\m_{KCl}=0,5\cdot74,5=37,25\left(g\right)\\m_{ddHCl}=\dfrac{0,5\cdot36,5}{10,95\%}\approx166,67\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{SO_2}=0,25\cdot64=16\left(g\right)\)

\(\Rightarrow m_{dd}=m_{K_2SO_3}+m_{ddHCl}-m_{SO_2}=190,17\left(g\right)\) \(\Rightarrow C\%_{KCl}=\dfrac{37,25}{190,17}\cdot100\%\approx19,59\%\)

d) PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Theo PTHH: \(n_{NaOH}=n_{HCl}=0,5\left(mol\right)\) \(\Rightarrow V_{NaOH}=\dfrac{0,5}{0,5}=1\left(l\right)\)

`n_[Fe_2 O_3]=20/160=0,125(mol)`

`n_[HCl]=[10,95.300]/[100.36,5]=0,9(mol)`

`Fe_2 O_3 +6HCl->2FeCl_3 +3H_2 O`

  `0,125`        `0,75`       `0,25`                          `(mol)`

Ta có: `[0,125]/1 < [0,9]/6 =>HCl` dư, `Fe_2 O_3` hết

   `C%_[FeCl_3]=[0,25.162,5]/[20+300].100=12,7%`

   `C%_[HCl(dư)]=[(0,9-0,75).36,5]/[20+300].100=1,7%`