Bài 1 : 

$R + 2HCl \to RCl_2 + H_2$
n R = n H2 = 2,24/22,4 = 0,1(mol)

M R = 2,4/0,1 = 24(Mg) - Magie

Bài 2 : 

$2R + 6HCl \to 2RCl_3 + 3H_2$
n H2 = 3,36/22,4 = 0,15(mol)

n R = 2/3 n H2 = 0,1(mol)

M R = 2,7/0,1 = 27(Al) - Nhôm

a) \(PTHH:2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\)

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

\(n_{Al}=\dfrac{0,15.2}{3}=0,1\left(mol\right)\)

\(m_{Al}=0,1.27=2,7\left(g\right)\)

c)\(n_{AlCl_3}=\dfrac{0,15.2}{3}=0,1\left(mol\right)\)

\(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)

pls help me mng ơi!Tuần sau mnh thi mất r!T_T

a, 2Al + 6HCl -> 2AlCl₃ + 3H₂  (1)

b, n_H2=\(\dfrac{V}{22.4}\)=\(\dfrac{3,36}{22,4}\)=0,15 (mol)

Từ pt (1)

-> n_Al=\(\dfrac{2}{3}\).n_H2 = \(\dfrac{2}{3}\).0,15 = 0,1 (mol)

-> m_Al = n.M = 0,1 . 27 = 2,7 (g)

c, Từ pt (1)

-> n_AlCl3= \(\dfrac{2}{3}\).n_H2 = \(\dfrac{2}{3}\). 0,15 = 0,1 (mol)

-> m_AlCl3= n.M = 0,1 .133.5 = 13,35 (g)

Giả sử KL có hóa trị n.

PT: \(2M+2nHCl\rightarrow2MCl_n+nH_2\)

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_M=\dfrac{2}{n}nH_2=\dfrac{0,4}{n}\left(mol\right)\)

\(\Rightarrow M_M=\dfrac{4,8}{\dfrac{0,4}{n}}=12n\left(g/mol\right)\)

Với n = 2 thì M_M = 24 (g/mol) là tm

Vậy: M là Mg.

Ta có: \(n_{MgCl_2}=n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgCl_2}=0,2.95=19\left(g\right)\)

1. Gọi chung hh 2 KL trên là A.

BT e, có: n_A = 2n_H2 ⇒ n_H2 = 0,05 (mol)

⇒ V_H2 = 0,05.22,4 = 1,12 (l)

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

BT e, có: 3n_Al = 2n_H2 ⇒ n_Al = 0,1 (mol)

\(\Rightarrow m_{Al_2O_3}=7,8-m_{Al}=5,1\left(g\right)\)

\(\Rightarrow\%m_{Al_2O_3}=\dfrac{5,1}{7,8}.100\%\approx65,38\%\)

a)

\(PTHH:2Al+6HCl->2AlCl_3+3H_2\)

                 1,3<---4<-------1,3<---------2                      

b)

\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{44,8}{22,4}=2\left(mol\right)\)

\(m_{AlCl_3}=n\cdot M=1,3\cdot\left(27+35,5\cdot3\right)=173,55\left(g\right)\)

\(m_{Al}=n\cdot M=1,3\cdot27=35,1\left(g\right)\)

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b, Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

nH₂=6.72/22,4=0,3(mol)
PTHH: 2Al +   6HCl --> 2AlCl₃ + 3H₂
bài ra:  0,2 <-- 0,6  <--   0,2   <-- 0,3  /mol
a) mHCl = 0,6.36,5=21,9(g)
b) mAl = 0.2.24 = 4,8(g)
 

\(PTHH:2Al+6HCl->2AlCl_3+3H_2\)

              0,2<--0,6<----------0,2<------0,3    (mol)

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

\(m_{HCl}=n\cdot M=0,6\cdot\left(1+35,5\right)=21,9\left(g\right)\)

\(m_{AlCl_3}=n\cdot M=0,2\cdot\left(27+35,5\cdot3\right)=26,7\left(g\right)\)

a, PT: 

Ta có: nAl=23nH2=0,2(mol)

⇒mAl=0,2.27=5,4(g)

a, PT: 

Ta có: nAl=23nH2=0,2(mol)

⇒mAl=0,2.27=5,4(g)