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)

$Mg + 2HCl \to MgCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2o$

b)

Theo PTHH : $n_{Mg} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$m_{Mg} = 0,2.24 = 4,8(gam)$

$m_{MgO} = m_{hh} - m_{Mg} = 12,8 - 4,8 = 8(gam)$

c)

$n_{MgO} = \dfrac{8}{40} = 0,2(mol)$
$n_{HCl} = 2n_{Mg} + 2n_{MgO} = 0,8(mol)$

$m_{dd\ HCl} = \dfrac{0,8.36,5}{14,6\%} = 200(gam)$

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

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

PTHH :

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

0,1      0,2            0,1      0,1

\(a,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(b,V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,2}{1}=0,2\left(l\right)\)

\(c,m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

Tên gọi : Sắt (II) Clorua

\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{Mg}=\dfrac{4,8}{24}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1       0,2            0,1         0,1

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

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

\(V_{H_2}=\left(0,1+0,2\right).22,4=6,72l\\ b)V_{ddHCl}=\dfrac{0,2+0,4}{2}=0,3l\\ c)m_{muối}=0,1.127+95.0,2=31,7g\)

a) \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

Theo PTHH ta có: \(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\)

\(\Rightarrow m_{MgCl_2}=n_{MgCl_2}.M_{MgCl_2}=0,1.95=9,5\left(g\right)\)

b) Theo PTHH ta có: \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,1.22,4=2,24\left(l\right)\)

c) \(2H_2+O_2\rightarrow2H_2O\)

Theo PTHH: \(n_{H_2O}=\dfrac{0,1.2}{2}=0,1\left(mol\right)\)

\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=0,1.18=1,8\left(g\right)\)

a)mMg= 2,4/24=0,1(mol) nMgCl2=0,1.1/1=0,1 (mol) mMgCl2= 0,1.95=9,5(g) b)nH2=0,1.1/1=0,1(mol) v=n.22,4=2,24(lít

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:

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

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

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

giúp mik gấp vs ạ mik cảm ơn

\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\ a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b,Theo.PTHH:n_{H_2}=n_{Mg}=0,3\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,3\cdot22,4=6,72\left(l\right)\)

Ta có sau p/ứ muối tạo thành là \(MgCl_2\)

Do đó \(m=m_{MgCl_2}=0,3\cdot95=28,5\left(g\right)\)