Theo gt ta có: $n_{Al}=0,2(mol)$

$2Al+6HCl\rightarrow 2AlCl_3+3H_2$

a, Ta có: $n_{H_2}=0,3(mol)\Rightarrow V_{H_2}=6,72(l)$

b, Ta có: $n_{HCl}=0,6(mol)\Rightarrow \%C_{HCl}=10,95\%$

c, Sau phản ứng dung dịch chứa 0,2 mol AlCl3

Suy ra $\%C_{AlCl_3}=13,03\%$

a)

\(Zn + 2HCl \to ZnCl_2 + H_2\)

b)

\(n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\)

Theo PTHH : \(n_{HCl} = 2n_{H_2} = 0,1.2 = 0,2(mol)\)

c)

Ta có :

 \(n_{ZnCl_2} = n_{H_2} = 0,1(mol)\\ \Rightarrow m_{ZnCl_2} = 0,1.136 = 13.6(gam)\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,1--------------->0,1---->0,1

=> m_FeCl2  = 0,1.127 = 12,7(g)

c) V_H2 = 0,1.22,4 = 2,24(l)

\(n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ a,Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{FeCl_2}=n_{H_2}=0,1(mol)\\ a,m_{FeCl_2}=0,1.127=12,7(g)\\ b,V_{H_2}=0,1.22,4=2,24(l)\)

56 là nguyên tử khối cùa Fe nhé , em có thể xem lại trong bảng.

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

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

\(0.2.......0.4...................0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(C_{M_{HCl}}=\dfrac{0.4}{0.4}=1\left(M\right)\)

\(n_{CuO}=\dfrac{8}{80}=0.1\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(1............1\)

\(0.1.........0.2\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.2}{1}\Rightarrow H_2dư\)

\(n_{Cu}=n_{CuO}=0.1\left(mol\right)\)

\(m_{Cu}=0.1\cdot64=6.4\left(g\right)\)

Chúc em học tốt và có những trải nghiệm tuyệt vời tại hoc24.vn nhé !

a) nFe=0,2(mol)

PTHH: Fe +  2HCl -> FeCl2 + H2

0,2_________0,4____0,2___0,2(mol)

V(H2,dktc)=0,2.22,4=4,48(l)

b) VddHCl=0,4/0,4=1(l)

c) nCuO=0,1(mol)

PTHH: CuO + H2 -to-> Cu + H2O

Ta có: 0,2/1 > 0,1/1

=> CuO hết, H2 dư, tính theo nCuO

-> nCu=nCuO=0,1(mol)

=>mCu=0,1.64=6,4(g)

CMHCl=0.40.4=1(M)CMHCl=0.40.4=1(M)

LTL:0.11<0.21⇒H2dưLTL:0.11<0.21⇒H2dư

nCu=nCuO=0.1(mol)nCu=nCuO=0.1(mol)

mCu=0.1⋅64=6.4(g)

a) \(Ca+2H_2O-->Ca\left(OH\right)_2+H_2\)

\(n_{Ca}=\dfrac{5,6}{40}=0,14\left(mol\right)\)

b)Theo PT:

\(n_{Ca\left(OH\right)_2}=n_{Ca}=0,14\left(mol\right)\)

=> \(C_M=0,14.0,2=0,028\left(M\right)\)

PTHH: \(Ca\left(OH\right)_2+CO_2-->CaCO_3+H_2O\)

Theo PT:

\(n_{CO_2}=n_{Ca\left(OH\right)_2}=0,14\left(mol\right)\)

=> \(V_{CO_2}=0,14.22,4=3,136\left(lit\right)\)

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%