HCl+NaOH->NaCl+H2O

0,2-------0,2----0,2

n HCl=0,2 mol

=>m NaOH=0,2.40=8g

->mdd NaOH=40g

=>m NaCl=0,2.58,5=11,7g

\(NaOH+HCl\rightarrow NaCl+H_2O\)

    0,2        0,2           0,2              ( mol )

\(n_{HCl}=1.0,2=0,2mol\)

\(m_{NaOH}=0,2.40:20\%=40g\)

\(m_{NaCl}=0,2.58,5=11,7g\)

\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3\downarrow+3NaCl\\ 2NH_3+\dfrac{3}{2}O_2\underrightarrow{t^o}N_2+3H_2O\)

\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)

\(2NH_3+\dfrac{3}{2}O_2\rightarrow\left(t^o\right)N_2+3H_2O\)

\(n_S=\dfrac{3,2}{32}=0,1mol\)

\(S+O_2\rightarrow\left(t^o\right)SO_2\)

0,1 0,1             0,1   ( mol )

Tỉ lệ : 1:1:1

\(V_{O_2}=V_{SO_2}=0,1.22,4=2,24l\)

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\\ PTHH:S+O_2\underrightarrow{t^o}SO_2\\ Mol:0,1\rightarrow0,1\rightarrow0,1\\ V_{SO_2}=V_{O_2}=0,1.22,4=2,24\left(l\right)\)

   fe2o3 +3h2so4 -> fe2(so)4 + 3h2o

  0,03125-----------------0,015625 mol

n Fe2O3=\(\dfrac{5}{160}\)=0,03125 mol

=>m Fe2(SO4)3=0,015625.400=6,25g

Zn+2HCl->ZnCl2+H2

0,2----------------------0,2

n Zn=\(\dfrac{13}{65}\)=0,2 mol

=>VH2=0,2.22,4=4,48l

\(n_{H_2SO_4}=\dfrac{150.9,8\%}{98}=0,15\left(mol\right)\\ H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+H_2O+CO_2\\ n_{Na_2CO_3}=n_{H_2SO_4}=0,15\left(mol\right)\\ \Rightarrow m_{ddNa_2CO_3}=\dfrac{0,15.106}{10,6\%}=150\left(g\right)\\ n_{CO_2}=n_{H_2SO_4}=0,15\left(mol\right)\\ m_{ddsaupu}=150+150-0,15.44=293,4\left(g\right)\\ n_{Na_2SO_4}=n_{H_2SO_4}=0,15\left(mol\right)\\ C\%_{Na_2SO_4}=\dfrac{0,15.142}{293,4}.100=7,26\%\)

chị ơi cho em hỏi tại sao lại 150* 9,8% lại chia cho 98 ạ

4

4P+5O2-to>2P2O5

0,2---0,25 -----0,1 mol

n P=0,2 mol

=>m P2O5=0,1.142=14,2g

=>Vkk=0,25.4.22,4=28l

Bài 5

3Fe+2O2-to>Fe3O4

0,3----0,2-----0,1 mol

n Fe3O4=0,1 mol

=>m Fe=0,3.56=16,8g

=>VO2=0,2.22,4=4,48l

2KMnO4-to>K2MnO4+MnO2+O2

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

=>m KMnO4=0,4.158=63,2g

a)

(1) BaCO₃ --t^o--> BaO + CO₂

(2) BaO + H₂O --> Ba(OH)₂

(3) Ba(OH)₂ + CO₂ --> BaCO₃ + H₂O

b)

(1) 3Fe + 2O₂ --t^o--> Fe₃O₄

(2) Fe₃O₄ + 4H₂ --t^o--> 3Fe + 4H₂O

b)3Fe+2O2->Fe3O4;

4CO+Fe3O4->3Fe+4CO2

BaCO3->BaO+CO2

;BaO+H2O->Ba(OH)2;

Ba(OH)2+CO2->BaCO3+H2O

Công suất của nó là

\(P=F.v=400.5=2000\left(W\right)\\ \Rightarrow A\)

A

\(m_{H_2SO_4}=150.9,8\%=14,7\left(g\right)\\ n_{H_2SO_4}=\dfrac{14,7}{98}=0,3\left(mol\right)\\ PTHH:H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2\uparrow+H_2O\\ Mol:0,3\rightarrow0,3\rightarrow0,3\rightarrow0,3\)

\(m_{Na_2CO_3}=0,3.106=31,8\left(g\right)\\ m_{ddNa_2CO_3}=\dfrac{31,8}{10,6\%}=300\left(g\right)\\ m_{Na_2SO_4}=0,3.142=42,6\left(g\right)\\ m_{CO_2}=0,3.44=13,2\left(g\right)\\ m_{dd}=150+300-13,2=436,8\left(g\right)\\ C\%_{Na_2SO_4}=\dfrac{42,6}{436,8}=9,75\%\)

mH2SO4 =mdd H2SO4.C% : 100% = 400.9,8% :100% = 39,2 (g)

=> nH2SO4 = mH2SO4 : MH2SO4 = 39,2: 98 = 0,4 (mol)

PTHH: H2SO4 + Na2CO3 ---> Na2SO4 + CO2 + H2O

0,4 ---->0,4 -----------> 0,4 -------> 0,4 (mol)

a) Theo PTHH: nNa2CO3 = nH2SO4 = 0,4 (mol)

=> mNa2CO3 = nNa2CO3. MNa2CO3 = 0,4.106 = 42,4 (g)

=> mdd Na2CO3 = mNa2CO3. 100% : C% = 42,4.100% : 10% = 424 (g)

b) Theo PTHH: nCO2 = nH2SO4 = 0,4 (mol)

=> VCO2(đktc) = 0,4.22,4 = 8,96 (lít)

c) Theo PTHH: nNa2SO4 = nH2SO4 = 0,4 (mol)

=> mNa2SO4 = nNa2SO4. MNa2SO4 = 0,4.142 = 56,8 (g)

mdd A = mdd H2SO4 + mdd Na2CO3 = 400 + 424 = 824 (g)

dd A chứa Na2SO4

=> C% Na2SO4 = (mNa2SO4 : mddA).100% = (56,8 : 824).100% = 6,89%