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

nFe = 5.6/56 = 0.1 (mol) 

nHCl = 0.2*2 = 0.4 (mol) 

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

LTL : 0.1/1 < 0.4/2 => HCl dư 

mHCl dư = ( 0.4 - 0.2 ) * 36.5 = 7.3 (g) 

VH2 = 0.2*22.4 = 4.48 (l) 

CM FeCl2 = 0.1/0.2 = 0.5(M)

CM HCl dư = 0.2 / 0.2 = 1(M) 

n Al=0,25 mol

2Al + 6HCl → 2AlCl₃ + 3H₂

0,25---0,75------0,25------0,375 mol

=>VH2=0,375.22,4=8,4l

=>m AlCl3=0,25.133,5=33,375g

=>CM HCl=\(\dfrac{0,75}{2}\)=0,375M

\(1,n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

           0,3---->0,6------------------>0,3

\(2,C_{M\left(HCl\right)}=\dfrac{0,6}{0,3}=2M\\ 3,V_{H_2}=0,3.22,4=6,72\left(l\right)\)

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

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\\C_{M_{FeCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)