3Fe + 2O2 => Fe3O4

m O2 pư = m oxit - mFe = 14 - 9 = 5 (g)

a) = 36,4x9+36,4x1

=36,4x(9+1)

=36,4x10

=364

b)  (8,27-7,27)+(7,16-6,16)+(9,33-8,33)

=1+1+1

=3

bài 4

nP2O5 = 0,1 nNaOH = 0,4mol

cho P2O5 vào dd

P2O5 => 2H3PO4

0,1--------> 0,2

ta có \(\frac{nNaOH}{nH3PO4}\) = 2 => tạo Na2HPO4

2NaOH + H3PO4 => Na2HPO4 + 2H2O

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

mdd = 14,2 + 200 = 214,2

=> C% Na2HPO4 = \(\frac{0,2.142}{214,2}\) .100% = 13,258%

2P => P2O5

0,2----> 0,1

P2O5 + 4NaOH => 2Na2HPO4 + H2O

0,1-----> 0,4

=> mdd NaOH = \(\frac{0,4.40}{32\%}\) = 50 (g)

bài 2

coi P2O5 là dd

=> trong P2O5 C% H3PO4 = \(\frac{98.2}{142}\) = 138%

142 138 X- 23,72

X

500 23,72 138- X

=> \(\frac{X-23,72}{138-X}\) = \(\frac{142}{500}\) => X = 49%

bài 1

nH3PO4 = 0,05

3NaOH + H3PO4 => Na3PO4 +3 H2O

0,15 <-------0,05

=> V NaOH = 0,15/1=0,15l - 150ml

4Na + O2 => 2Na2O

Na2O + H2O => 2NaOH

NaOH + NaHSO4 => Na2SO4 + H2O

NaOH + NaHCO3 => Na2CO3 + H2O

2NaOH + (NH4)2SO4 => Na2SO4 + 2NH3 + 2H2O

3NaOH + AlCl3 => 3NaCl + Al(OH)3

NaOH + Al(OH)3 => NaAlO2 + 2H2O

nNaOH = 6/40 = 0,15

=> CM NaOH = 0,15/ 1,5 = 0,1M

nKOH = 0,075 nMgCl2 = 0,05 => MgCl2 dư

2KOH + MgCl2 => 2KCl + Mg(OH)2

=> n Mg(OH)2 = 1/2nKOH = 0,0375

=> m Mg(OH)2 = 2,175 (g)

c% MgCl2 dư = 0,95%

C% KCl = 4,47%