a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

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

\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{20\%}=36,5\left(g\right)\)

c, \(n_{MgCl_2}=n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 2,4 + 36,5 - 0,1.2 = 38,7 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,1.95}{38,7}.100\%\approx24,55\%\)

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

b) \(n_{Mg}=\dfrac{4,8}{24}=0,2mol\) 

\(n_{HCl}=2.n_{Mg}=0,2.2=0,4mol\)

\(\Rightarrow m_{HCl}=n.M=0,4.36,5=14,6g\)

c) \(n_{H_2}=n_{Mg}=0,2mol\) 

Thể tích khí hidro sinh ra (ở đktc): 

\(V_{H_2}=0,2.24,79=4,958l.\)

bà định cân hết các box hay sao v:))

mHCl= 3,65%.400= 14,6(g) => nHCl=14,6/36,5=0,4(mol)

a) PTHH: Mg +2 HCl -> MgCl2 + H2

0,2__________0,4_____0,2____0,2(mol)

V(H2,đktc)=0,2.22,4=4,48(l)

b)mMg=0,2.24=4,8(g)

c) mMgCl2= 0,2.95=19(g)

mddMgCl2= 400+4,8 - 0,2.2= 404,4(g)

=> C%ddMgCl2= (19/404,4).100=4,698%

\(n_{HCl}=\dfrac{18.25}{36.5}=0.5\left(mol\right)\)

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

\(b.\)

\(n_{Mg}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.5=0.25\left(mol\right)\)

\(m_{Mg}=0.25\cdot24=6\left(g\right)\)

\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)

\(c.\)

\(V_{H_2\left(tt\right)}=5.6\cdot90\%=5.04\left(l\right)\)

`a)PTHH:`

`2Al + 6HCl -> 2AlCl_3 + 3H_2`

`0,2`     `0,6`                               `0,3`       `(mol)`

`n_[Al]=[5,4]/27=0,2(mol)`

`b)V_[H_2]=0,3.22,4=6,72(l)`

`c)m_[dd HCl]=[0,6.36,5]/10 . 100 =219(g)`

n_Al  = \(\dfrac{5,4}{27}\) = 0,2 (mol)

 2Al + 6HCl -> 2AlCl₃ + 3H₂

     0,2     0,6       0,2      0,3

=>VH2=0,3.22,4=6,72l

=>mHcl=0,6.36,5=21,9g

=>mdd=219g

nMg = 3,6 : 24 = 0,15 (mol) 
pthh : Mg + 2HCl --> MgCl2 + H2 
           0,15-----------> 0,15 --->0,15 (mol) 
mMgCl2 = 0,15 . 95 = 14,25 (mol) 
VH2 (đkc)= 0,15. 24,79 = 3,718(l) 
 

\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\)

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

0,15      0,3         0,15       0,15

\(m_{MgCl_2}=0,15\cdot95=14,25g\)

\(V_{H_2}=0,15\cdot22,4=3,36l\)

\(nHCl=0,2.0,3=0,06\\ 2Al+6HCl=>2AlCl3+3H2\\ =>nAl=0,02\left(mol\right)\\ =>mAl=0,02.27=0,54\left(g\right)\\ tacónAlCl3=0,02\left(mol\right)\\ =>Cm\left(AlCl3\right)=\dfrac{0,02}{0,2}=0,1\left(M\right)\)

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

b, \(n_{HCl}=0,2.0,3=0,06\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{1}{3}n_{HCl}=0,02\left(mol\right)\)

\(\Rightarrow m_{Al}=0,02.27=0,54\left(g\right)\)

c, \(n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,02\left(mol\right)\)

\(\Rightarrow C_{M_{AlCl_3}}=\dfrac{0,02}{0,2}=0,1\left(M\right)\)

a)

 \(PTHH:Mg+2HCl->MgCl_2+H_2\)

               2<------4<----------2<---------2   (mol)

b)

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

\(m_{HCl}=n\cdot M=4\cdot\left(1+35,5\right)=146\left(g\right)\)

c)

\(m_{MgCl_2}=n\cdot M=2\cdot\left(24+71\right)=190\left(g\right)\)

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

\(\text{a)}Mg+2HCl\rightarrow MgCl_2+H_2\)

             \(2mol\)       \(1mol\)       \(1mol\)

              \(4mol\)       \(2mol\)       \(2mol\)

\(b)m_{HCl}=n.M=4.36,5=146\left(g\right)\)

\(c)m_{MgCl_2}=n.M=2.95=190\left(g\right)\)