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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

____0,15<--0,3<-------------0,15

=> m_Fe = 0,15.56 = 8,4 (g)

b) \(C_{M\left(ddHCl\right)}=\dfrac{0,3}{0,05}=6M\)

\(a)\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\)

\(b)\ n_{H_2} = \dfrac{1,12}{22,4} = 0,05(mol)\\ n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{0,1}{3}(mol)\\ \Rightarrow m_{Al} = \dfrac{0,1}{3}.27= 0,9\ gam\)

\(c)\ n_{HCl} = 2n_{H_2} = 0,05.2 = 0,1(mol)\\ \Rightarrow C_{M_{HCl}} = \dfrac{0,1}{0,2} = 0,5M\)

a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

b) Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(\Rightarrow n_{Al}=\dfrac{1}{30}\left(mol\right)\) \(\Rightarrow m_{Al}=\dfrac{1}{30}\cdot27=0,9\left(g\right)\)

b) Theo PTHH: \(n_{HCl}=2n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

\(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

a)

PTHH:

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

0,15<-0,3<-------------0,15

b)

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

c)

\(CM_{HCl}=\dfrac{0,3}{0,05}=6M\)

\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

a) Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

            1         2             1          1

          0,15     0,3                      0,15

b) \(n_{Fe}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

⇒ \(m_{Fe}=0,15.56=8,4\left(g\right)\)

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

50ml = 0,05l

\(C_{M_{ddHCl}}=\dfrac{0,3}{0,05}=6\left(M\right)\)

 Chúc bạn học tốt

Fe+2HCl->FeCl2+H2

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

n H2=0,3 mol

=>m Fe=0,3.56=16.8g

=>CM=0,6\0,1=6M

CuO+H2-tO>Cu+H2O

0.3---------------0,3

=>m Cu=0,3.64=19,2g

câu 2

FE+H2SO4->FeSO4+H2

0,25----0,25----0,25---0,25

n Fe=0,25 mol

=>mFeCl2=0,25.127=31,75g

=>VH2=0,25.22,4=5,6l

=>C% muối=\(\dfrac{31,75}{14+400-0,25.2}100=7,67\%\)

\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

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

\(0.15......0.3..................0.15\)

\(m_{Fe}=0.15\cdot56=8.4\left(g\right)\)

\(C_{M_{HCl}}=\dfrac{0.3}{0.05}=6\left(M\right)\)

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

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

Theo PT: \(n_{Zn}=n_{H_2SO_4}=n_{ZnCl_2}=n_{H_2}=0,15\left(mol\right)\)

b, m_Zn = 0,15.65 = 9,75 (g)

c, C_{M (H2SO4)} = 0,15/0,05 = 3 M

d, m_ZnSO4 = 0,15.161 = 24,15 (g)

Bạn tham khảo nhé!