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

PTHH: CaO + 2HCl → CaCl₂ + H₂O

Mol:     0,1                       0,1

\(\Rightarrow m_{CaCl_2}=0,1.111=11,1\left(g\right)\)

\(a.CaO+2HCl\rightarrow CaCl_2+H_2O\\ n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{CaCl_2}=n_{CaO}=0,2\left(mol\right)\\ m_{CaCl_2}=111.0,2=22,2\left(g\right)\)

Tên muối: Canxi clorua

\(b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\)

Giúp mình với mn ơi 😭

nH2 = 0,336/22,4 = 0,015 (mol)

PTHH: 2A + 6HCl -> 2ACl3 + 3H2

nACl3 = nA = 0,015 : 3 . 2 = 0,01 (mol)

M(A) = 0,27/0,01 = 27 (g/mol)

=> A là Al

mAlCl3 = 0,01 . 133,5 = 1,335 (g)

Câu 6.

\(n_{H_2}=\dfrac{0,336}{22,4}=0,015mol\)

\(2A+6HCl\rightarrow2ACl_3+3H_2\)

0,01                                0,015

\(\overline{M_A}=\dfrac{0,27}{0,01}=27đvC\)

\(\Rightarrow A\) là Al nhôm.

\(m_{AlCl_3}=0,01\cdot133,5=1,335g\)

a, Dấu hiệu xảy ra phản ứng là có khí thoát ra nè

b) CTHH của muối: ZnCl₂

PTHH: Zn +2 HCl -> ZnCl₂ + H₂

c) Áp dụng ĐLBTKL, ta có:

\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ \Leftrightarrow65+73=m_{ZnCl_2}+2\\ \Leftrightarrow m_{ZnCl_2}=136\left(g\right)\)

\(n_{HCl}=\dfrac{73.4\%}{36,5}=0,08\left(mol\right)\\ MgO+2HCl\rightarrow MgCl_2+H_2O\\ 0,04........0,08.......0,04......0,04\left(mol\right)\\ m_{MgO}=0,04.40=1,6\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,04.95}{1,6+73-0,04.2}.100\approx5,099\%\)

\(m_{ct}=\dfrac{4.73}{100}=2,92\left(g\right)\)

\(n_{HCl}=\dfrac{2,92}{36,5}=0,08\left(mol\right)\)

Pt : \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

         1            2              1             1

       0,04       0,08         0,04

\(n_{MgO}=\dfrac{0,08.1}{2}=0,04\left(mol\right)\)

⇒ \(m_{MgO}=0,04.40=1,6\left(g\right)\)
\(n_{MgCl2}=\dfrac{0,08.1}{2}=0,04\left(mol\right)\)

⇒ \(m_{MgCl2}=0,04.95=3,8\left(g\right)\)

\(m_{ddspu}=1,6+73=74,6\left(g\right)\)

\(C_{MgCl2}=\dfrac{3,8.100}{74,6}=5,09\)⁰/₀

 Chúc bạn học tốt

Câu 1: 

\(Ca+Cl_2\rightarrow CaCl_2\)

\(n_{Cl_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

=>n_Ca=0,4(mol)

\(m_{Ca}=0.4\cdot12=4.8\left(g\right)\)

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) 

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