Câu 3 : 

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

            1            2             1          1

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

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

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

\(m_{ddHCl}=\dfrac{14,6.100}{10,95}=133,3\left(g\right)\)

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

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

d) \(n_{MgCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{MgCl2}=0,2.95=19\left(g\right)\)

\(m_{ddspu}=4,8+133,3-\left(0,2.2\right)=137,7\left(g\right)\)

\(C_{MgCl2}=\dfrac{19.100}{137,7}=13,8\)⁰/₀

 Chúc bạn học tốt

\(a)Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{Mg}=\dfrac{5,28}{24}=0,22mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,22...0,44........0,22........0,22\\ V_{H_2\left(đkc\right)}=0,22.24,79=5,4538l\\ b)C_{\%HCl}=\dfrac{0,44.26,5}{200}\cdot100=8,03\%\\ c)C_{\%MgCl_2}=\dfrac{0,22.95}{200+5,28}\cdot100\approx10,18\%\)

a) $CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
Theo PTHH : 

$n_{CO_2} = n_{CaCO_3} = \dfrac{10}{100} = 0,1(mol)$
$V_{CO_2} = 0,1.22,4 = 2,24(lít)$
b) $n_{HCl} = 2n_{CaCO_3} = 0,2(mol)$
$C_{M_{HCl}} = \dfrac{0,2}{0,25} = 0,8M$

c) $CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,1(mol)$
$m_{CaCO_3} = 0,1.100 = 10(gam)$

Mg+2HCl->MgCl2+H2

0,05--0,1------0,05-------0,05

n Mg=1,2\24=0,05 mol

=>VH2=0,05.22,4=1,12l

C) mình chưa hiểu đề bài do viết ko dấu

            Mg+    2HCl→       MgCl₂+    H₂

(mol)   0,1          0,2            0,1          0,1                                             

a) \(n_{Mg}=\dfrac{m}{M}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

→\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(lít\right)\)

b) Đổi: 100ml=0,1 lít

\(C_{M_{HCl}}=\dfrac{n}{V}=\dfrac{0,2}{0,1}=2M\)

c) \(m_{MgCl_2}=n.M=0,1.95=9,5\left(g\right)\)

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

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

______0,2---->0,3------------>0,1------>0,3______(mol)

=> V_H2 = 0,3.22,4= 6,72(l)

b) \(C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,1}=3M\)

\(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,1}{0,1}=1M\)

Câu 3:

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=n_{H_2}=0,3(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,3.24}{15,2}.100\%=47,37\%\\ \Rightarrow \%_{MgO}=100\%-47,37\%=52,63\%\)

\(n_{MgO}=\dfrac{15,2-0,3.24}{40}=0,2(mol)\\ \Rightarrow \Sigma n_{HCl}=0,3.2+0,2.2=1(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{1.36,5}{10\%}=365(g)\\ \Sigma n_{MgCl_2}=0,2+0,3=0,5(mol)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{0,5.95}{15,2+365}.100\%=12,49\%\)

\(PTHH:Mg+2H_2SO_{4(đ)}\to MgSO_4+2H_2O+SO_2\uparrow\\ MgO+H_2SO_4\to MgSO_4+H_2O\\ \Rightarrow n_{SO_2}=n_{Mg}=0,3(mol)\\ \Rightarrow V_{SO_2}=0,3.22,4=6,72(l)\)

nH2=\(\frac{6,72}{22,4}=0,3\)mol

PTHH

         M+2HCl--> MCl₂+H₂

      0,3mol<---------------0,3mol

=>M_M=\(\frac{19,5}{0,3}=64\)

=> km loại là kẽm (Zn)

b) n_NaOH=0,2.1=0,2 mol

PTHH

         NaOH+HCl-->NaCl + H2O

         0,2 mol--> 0,2 mol

---> thể tích HCl 1M đã dùng là V=\(\frac{0,2+0,3}{1}=0,5\)lít

=> C_M(ZnCl2)=\(\frac{0,3}{0,5}=0,6M\)