a) \(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

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

Số mol : 2mol.........6mol

Sau p/ư : 0,2mol........ x

\(n_{HCl}=x=\dfrac{0,2.6}{2}=0,6\left(mol\right)\)

Ta có : \(n_{HCl}=\dfrac{16\%.m_{HCl}}{100\%.36,5}\)

\(\Rightarrow0,6=\dfrac{16.m_{HCl}}{100.36,5}\)

\(\Rightarrow m_{HCl}=\dfrac{2190}{16}=136,875\left(g\right)\)

b) \(n_{AlCl_3}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

\(m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)

\(\Rightarrow C_{\%ddAlCl_3}=\dfrac{40,05}{5,4+136,875}.100\approx27,95\%\)

nAl=5,4/27=0,2(mol)

PT:

2Al + 6HCl -> 2AlCl3 + 3H2↑

0.2........0.6........0.2..........0.3 (mol)

a) mHCl=0,6.36,5=21,9 (g)

b)m_AlCl3=0.2*133.5=26.7(g)

=> C% (AlCl₃)=(26.7/(21.9*100/16+5.4))*100%=18.766%

n_Al = 0,2 mol

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

0,2.....0,6...........0,6..........0,3

\(\Rightarrow\) m_{HCl đã dùng} = \(\dfrac{0,6.36,5.100}{16}\) = 136,875 (g)

\(\Rightarrow\) m_{dd sau phản ứng} = 5,4 + 136,875 - ( 0,3.2 ) = 141,675 (g)

\(\Rightarrow\) C%_{dd muối sau phản ứng} = \(\dfrac{0,2.133,5.100}{141,675}\) \(\approx\) 18,8%

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

Pt: 2Al + 6HCl --> 2AlCl₃ + 3H₂

0,2 mol-> 0,6 mol----------> 0,3 mol

V_H2 sinh ra = 0,3 . 22,4 = 6,72 (lít)

C_{M HCl} đã dùng = \(\dfrac{0,6}{0,2}=3M\)

n_CuO = \(\dfrac{12}{80}=0,15\) mol

Pt: CuO + H₂ --to--> Cu + H₂O

0,15 mol-------------> 0,15 mol

Xét tỉ lệ mol giữa CuO và H₂:

\(\dfrac{0,15}{1}< \dfrac{0,3}{1}\)

Vậy H₂ dư

m_Cu tạo thành = 0,15 . 64 = 9,6 (g)

Bài1

Fe +2HCl----> FeCl2 +H2

Ta có

m\(_{HCl}=\frac{14,66.200}{100}=29,32\left(g\right)\)

n\(_{HCl}=\frac{29,32}{36,5}=0,8\left(mol\right)\)

Theo pthh

n\(_{Fe}=\frac{1}{2}n_{HCl}=0,4\left(mol\right)\)

m=m\(_{Fe}=0,4.56=22,4\left(g\right)\)

Theo pthh

n\(_{FeCl2}=n_{Fe}=0,4\left(mol\right)\)

m\(_{FeCl2}=0,4.127=50,8\left(g\right)\)

m\(_{H2}=0,8\left(g\right)\)

mdd= 22,4+200=0,8=221,6(g)

C%=\(\frac{50,8}{221,6}.100\%=22,92\%\)

Chúc bạn học tốt

Bài 2

2K+2H2O--->2KOH+H2

Ta có

n\(_K=\frac{15,6}{137}=0,1\left(mol\right)\)

Theo pthh

n\(_{H2}=\frac{1}{2}n_K=0,05\left(mol\right)\)

m\(_{H2}=0,1\left(g\right)\)

m=m\(_{H2O}=200+0,1-15,6=184,5\left(g\right)\)

V\(_{H2}=0,05.22,4=1,12\left(l\right)\)

Chúc bạn học tốt

nZn=13/65=0,2mol

đổi 200ml=0,2l

pt : Zn + 2 HCl -----> ZnCl2 + H2

npứ: 0,2----->0,4---------->0,2

CM(HCl)=0,4/0,2=2M

CM(ZnCl2) = 0,2/0,2 = 1M

Bài 1:

Zn + 2HCl → ZnCl₂ + H₂

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Zn}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

Theo PT: \(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\)

\(\Rightarrow C_{M_{ZnCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Bài 2:

Gọi CTHH của oxit là Fe_xO_y

PTHH: Fe_xO_y + 2yHCl → xFeCl_{\(\dfrac{2y}{x}\)} + yH₂O

Ta có: \(n_{FeCl_{\dfrac{2y}{x}}}=\dfrac{12,7}{56+35,5\times\dfrac{2y}{x}}=\dfrac{12,7}{56+\dfrac{71y}{x}}\left(mol\right)\)

Theo PT: \(n_{Fe_xO_y}=\dfrac{1}{x}n_{FeCl_{\dfrac{2y}{x}}}=\dfrac{1}{x}\times\dfrac{12,7}{56+\dfrac{71y}{x}}=\dfrac{12,7}{56x+71y}\left(mol\right)\)

⇒ \(M_{Fe_xO_y}=\dfrac{m_{Fe_xO_y}}{n_{Fe_xO_y}}=7,2\div\dfrac{12,7}{56x+71y}=\dfrac{403,2x+511,2y}{12,7}\left(g\right)\)

Ta có: \(56x+16y=\dfrac{403,2x+511,2y}{12,7}\)

\(\Leftrightarrow711,2x+203,2y=403,2x+511,2y\)

\(\Leftrightarrow308x=308y\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{308}{308}=\dfrac{1}{1}\left(tốigiản\right)\)

Vậy \(x=1;y=1\)

Vậy CTHH của oxit sắt là FeO