a. \(Al+2HCl\rightarrow2AlCl_3+3H_2\)

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

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

bạn xem lại đề

a) \(m_{CO_2}=2.44=88\left(g\right)\)

b)\(n_{O_2}=\dfrac{32}{32}=1\left(mol\right)\Rightarrow V_{H_2}=1.24,79=24,79\left(l\right)\)

\(a.m_{CO_2}=2.44=88\left(g\right)\\ b.n_{O_2}=\dfrac{32}{32}=1\left(mol\right)\\ V_{O_2\left(đktc\right)}=1.22,4=22,4\left(l\right)\)

a,\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     0,1    0,2                             0,1

b,\(m_{Fe}=0,1.56=5,6\left(g\right)\)

\(\Rightarrow\%m_{Fe}=\dfrac{5,6.100\%}{12}=46,67\%;\%m_{Cu}=100-46,67=53,33\%\)

c,\(m_{HCl}=0,2.36,5=7,3\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{7,3.100}{14,6}=50\left(g\right)\)

a) tỉ khối của các khí với H₂

\(d_{\dfrac{O_2}{H}}=32\) , \(d_{\dfrac{N_2}{H}}=28,d_{\dfrac{CO_2}{H}}=44,d_{\dfrac{O_3}{H}}=48,d_{\dfrac{SO2}{H}}=64,d_{\dfrac{H_2}{H}}=2\)

b) So với không khí

\(d_{\dfrac{O_2}{kk}}=\dfrac{32}{29}=1,1,d_{\dfrac{N_2}{kk}}=\dfrac{28}{29}=0,9655,d_{\dfrac{CO2}{kk}}=\dfrac{44}{29}=1,52,d_{\dfrac{O3}{kk}}=\dfrac{48}{29}=1,655,d_{\dfrac{SO2}{kk}}=\dfrac{64}{29}=2,21,d_{\dfrac{H2}{kk}}=\dfrac{2}{29}=0,069\)

Mik ko thấy gì cả

\(n_{Al}=\dfrac{16,2}{27}=0,6\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 0,6........0,9...........0,3........0,9\left(mol\right)\\ a.V_{H_2\left(đktc\right)}=0,9.22,4=20,16\left(l\right)\\ b.C_{MddH_2SO_4}=\dfrac{0,9}{0,5}=1,8\left(M\right)\\ c.C_{MddX}=C_{MddAl_2\left(SO_4\right)_3}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:     0,4                                         0,4

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

\(m_{hh}=22,4+5=27,4\left(g\right)\)

\(\%m_{Fe}=\dfrac{22,4.100\%}{27,4}=81,75\%;\%m_{Cu}=100-81,75=18,25\%\)

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

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,2     0,4                        0,2

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b,\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

c,\(m_{ddHCl}=\dfrac{14,6.100\%}{15\%}=97,33\left(g\right)\)

1/ MgO + 2HCl → MgCl₂ + H₂

2/ CO₂ + CaO → CaCO₃ 

3/ SO₂ +2NaOH → Na₂SO₃ + H₂O

4/ CaO + H₂O → Ca(OH)₂

5/ SO₃ + H₂O → H₂SO₄

3/

a,\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:    0,05    0,1                     0,1

\(V_{ddHCl}=\dfrac{0,1}{2}=0,05\left(l\right)\)

b,\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

4/

M₂ ở đâu vậy bn