\(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\\ a,n_{CH_3COOH}=0,2.1=0,2\left(mol\right)\Rightarrow n_{CO_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\Rightarrow V_{CO_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ b,CH_3COOH+C_2H_5OH⇌\left(H^+,t^o\right)CH_3COOC_2H_5+H_2O\\ n_{CH_3COOH}=\dfrac{50}{200}.0,2=0,05\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\\ Vì:\dfrac{0,5}{1}>\dfrac{0,05}{1}\Rightarrow Rượu.dư\\ \Rightarrow n_{este\left(LT\right)}=n_{axit}=0,05\left(mol\right)\\ \Rightarrow n_{este\left(TT\right)}=80\%.0,05=0,04\left(mol\right)\\ m_{CH_3COOC_2H_5}=88.0,04=3,52\left(g\right)\)

F e + 2 H C l → F e C L 2 + H 2

Vì H = 85% nên:

n H 2  = n F e   p ư = 0,01275 mol

V H 2 = 0,01275.22,4= 0,2856 lit

⇒ Chọn C.

Ca + 2H2O --> Ca(OH)2 + H2

0,2     0,4            0,2             0,2

nCa=8/40=0,2(mol)

b/

mCa(OH)2=0,2.74=14,8(g)

VH2=0,2.22,4.80%=3,584(l)

a) \(4Al+3O_2-^{t^o}\rightarrow2Al_2O_3\)

Bảo toàn khối lượng : \(m_{O_2}=12,24-8,1=4,14\left(g\right)\)

=>\(n_{O_2}=\dfrac{207}{1600}\left(mol\right)\)

Vì O₂ chiếm 20% thể tích không khí

 \(V_{kk}=\dfrac{\dfrac{207}{1600}.22,4}{20\%}=14,49\left(lít\right)\)

b) \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

Lập tỉ lệ : \(\dfrac{0,3}{4}>\dfrac{\dfrac{207}{1600}}{3}\)

=> Sau phản ứng Al dư

\(n_{Al\left(pứ\right)}=\dfrac{207}{1600}.\dfrac{4}{3}=0,1725\left(mol\right)\)

=> \(H=\dfrac{0,1725}{0,3}.100=57,5\%\)

c) D gồm Al2O3 và Al dư

\(n_{Al_2O_3}=\dfrac{2}{3}n_{O_2}=\dfrac{69}{800}\left(mol\right);n_{Al\left(dư\right)}=0,3-0,1725=0,1275\left(mol\right)\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

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

\(\Sigma n_{HCl}=\dfrac{69}{800}.6+0,1275.3=0,9\left(mol\right)\)

=> \(m_{HCl}=0,9.36,5=32,85\left(g\right)\)

a) \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

            0,4<--0,2-------->0,4

=> \(V_{H_2}=0,4.22,4=8,96\left(l\right)\)

b) \(m_{H_2O}=0,4.18=7,2\left(g\right)\)

c) m_HCl = 50 - 7,2 = 42,8 (g)

=> \(C\%_{ddHCl}=\dfrac{42,8}{50}.100\%=85,6\%\)

không bt sai đâu không nhưng nồng độ cao nhất của HCl ở khoảng 40% nhé :)

2H₂+O₂-to>2H₂O

0,4----0,2------0,4  mol

n _O2=\(\dfrac{4,48}{22,4}\)=0,2 mol

=>V_H2=0,4.22,4=8,96l

=>m _H2O=0,4.18=7,2g

c) ta có :mdd HCl =m H₂O+m HCl=50g

=>m _HCl=50-7,2=42,8g

=>C%_HCl=\(\dfrac{42,8}{50}.100\)=85,6%

Ta có: \(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\)

PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

______0,2_______0,4_____0,2____0,2 (mol)

a, \(m_{ddHCl}=\dfrac{0,4.36,5}{14,6\%}=100\left(g\right)\)

b, \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

c, m _{dd sau pư} = 20 + 100 - 0,2.44 = 111,2 (g)

\(\Rightarrow C\%_{CaCl_2}=\dfrac{0,2.111}{111,2}.100\%\approx19,96\%\)

\(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{H_2}=n_{Zn}=0,25\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

Zn + 2HCl -> ZnCl2 + H2

0.25                             0.25

nZn = 0.25 mol

\(V_{H2}=0.25\times22.4=5.6l\)

Zn + 2HCl \(\rightarrow\) ZnCl₂ + H₂

\(n_{Zn}=\dfrac{16,25}{65}=0,25mol\)

\(n_{H_2}=0,25mol\\ V_{H_2}=0,25.22,4=5,6l\)

1)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

____0,1----->0,15

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%=\dfrac{14,7}{250}.100\%=5,88\%\)

2)

\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

_______0,2------------------------------>0,2

=> V_CO2 = 0,2.22,4 = 4,48(l)

3)

\(n_A=\dfrac{18,4}{M_A}\left(mol\right)\)

PTHH: 2A + Cl₂ --t^o--> 2ACl

____\(\dfrac{18,4}{M_A}\)---------->\(\dfrac{18,4}{M_A}\)

=> \(\dfrac{18,4}{M_A}\left(M_A+35,5\right)=46,8=>M_A=23\left(Na\right)\)

4)

n_HCl = 0,2.3 = 0,6(mol)

PTHH: M + 2HCl --> MCl₂ + H₂

____0,3<-----0,6

=> \(M_M=\dfrac{7,2}{0,3}=24\left(Mg\right)\)