\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=0,1mol\\ m_{Zn}=0,1.65=6,5g\\ m_{Cu}=9,7-6,5=3,2g\\ b)C_{\%ZnCl_2}=\dfrac{0,1.136}{6,5+120-0,1.2}\cdot100=10,77\%\)

Gọi nMg = a (mol); nFe = b (mol)

24a + 56b = 10,4 (1)

nH2 = 6,72/22,4 = 0,3 (mol)

PTHH:

Mg + 2HCl -> MgCl2 + H2

a ---> a ---> a ---> a

Fe + 2HCl -> FeCl2 + H2

b ---> b ---> b ---> b

a + b = 0,3 (2)

(1)(2) => a = 0,2 (mol); b = 0,1 (mol)

mMg = 0,2 . 24 = 4,8 (g)

mFe = 0,1 . 56 = 5,6 (g)

a)PTHH Mg + 2HCl —> MgCl2 + H2

 Theo pt 1mol  2mol       1mol       1mol

Theo bài x mol                                   x mol

PTHH Fe +  2HCl —> FeCl2 + H2

Theo pt1mol  2mol       1mol       1mol

Theo bài 0,3-x                                   0,3-x

Số moi H2 (đktc)là nH2=V/22,4=6,72/22,4=0,3 moi

b)đặt H2 ở pt1 là x và H2 pt2 là 0,3-x

Ta được pt: 24x + (0,3-x)56 = 10,4g

=>giải pt ta được x=0,2

Khối lượng Mg là: mMg=n.m=24.0,2=4.8(g)

Khối lượng Fe là:mFe=n.m=(0,3-0,2).56=5,6(g)

a) PTHH : 

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

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

b) \(n_{H_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)

\(\Rightarrow n_{Mg}=n_{H_2}=0,3mol\)

\(m_{Mg}=n.M=0,3.24=7,2g\)

\(\Rightarrow m_{Fe}=m_{hh}-m_{Mg}=10,4-7,2=3,2g\)

a) 

Mg + 2HCl --> MgCl₂ + H₂

b)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

            0,3<-----------------0,3

=> m_Mg = 0,3.24 = 7,2 (g)

=> m_Ag = 10,4 - 7,2 = 3,2 (g)

c) \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{7,2}{10,4}.100\%=69,23\%\\\%m_{Ag}=\dfrac{3,2}{10,4}.100\%=30,77\%\end{matrix}\right.\)

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

b, Ta có: 24n_Mg + 56n_Fe = 10,4 (1)

Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(2\right)\)

Từ (1)  và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,2\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2.24=4,8\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)

c, Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\Rightarrow V_{ddHCl}=\dfrac{0,6}{0,5}=1,2\left(l\right)\)

a) Mg + 2HCl --> MgCl₂ + H₂

b) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

             0,3<--0,6<-----------0,3

=> m_Mg = 0,3.24 = 7,2 (g)

=> m_Ag = 10,4 - 7,2 = 3,2 (g)

c) \(V_{dd.HCl}=\dfrac{0,6}{0,5}=1,2\left(l\right)\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

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

0,3                                     0,3

\(m_{Mg}=0,3\cdot24=7,2g\Rightarrow m_{Ag}=10,4-7,2=3,2g\)

\(n_{HCl}=0,6mol\Rightarrow V_{HCl}=\dfrac{0,6}{0,5}=1,2l\)

a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,15.65=9,75\left(g\right)\)

\(\Rightarrow m_{Cu}=m_{hh}-m_{Zn}=21-9,75=11,25\left(g\right)\)

Bạn tham khảo nhé!

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, 24n_Mg + 27n_Al = 6,12 (1)

Theo PT: \(n_{H_2}=n_{Mg}+\dfrac{3}{2}n_{Al}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,12\left(mol\right)\\n_{Al}=0,12\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,12.24}{6,12}.100\%\approx47,06\%\\\%m_{Al}\approx52,94\%\end{matrix}\right.\)

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

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(n_{Mg}=n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(m_{Mg}=0.05\cdot24=1.2g\)

\(m_{MgO}=9.2-1.2=8\left(g\right)\)

a)

$Mg + 2HCl \to MgCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2O$
b)

Theo PTHH :

$n_{Mg} = n_{H_2} = \dfrac{1,12}{22,4} = 0,05(mol)$

$m_{Mg} = 0,05.24 = 1,2(gam)$
$m_{MgO} = 9,2 - 1,2 = 8(gam)$