Me—луг—нерозчинна у воді основа—оксид—сіль

1. m(S) = 32 g

N(S) - ?

n(S) = 32 / 32 = 1 mol

N(S) = 1 * 6.02*10^23 = 6.02*10^23 моль^-1 атомов.

2. m(H2SO4) = 98 g

n(H2SO4) - ?

n(H2SO4) = 98 / 98 = 1 mol

3. m(S) = 32 g

n(S) - ?

n(S) = 32 / 32 = 1 mol

4. n(S) = 2 mol

m(S) - ?

m(S) = 2 * 32 = 64 g

6. n(H2) = 3 mol

n(NH3) - ?

        3 mol  x mol

N2 + 3H2 = 2NH3

        3 mol  2 mol

3:3 = 2:x

3x = 6

x = 2

n(NH3) = 2 mol

7. n(H2SO4) = 3 mol

n(H2) - ?

3H2SO4 - следовательно 6 моль водорода.

n(H2) = 6 mol

8. m(C) = 100 g

n(C) - ?

n(C) = 100 / 12 = 8.3333 = 8.3 mol (округление)

1) 2Al + 6HCl = 2AlCl3 + 3H2

2Al + 6H + 6Cl = 2Al + 6Cl + 6H

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

Al + Cl3 + 3Na + 3OH = 3Na + 3Cl + Al(OH)3

Al + 3OH = Al(OH)3

3) 2Al(OH)3 = Al2O3 + 3H2O

2Al(OH)3 = 2Al + 3O + 3H2O

4) Al2O3 + 2NaOH + 3H2O = 2Na[Al(OH)4]

2Al + 3O + 2Na + 2OH + 3H2O = 2Na + 2Al(OH)4

2Al + 3O + 2OH + 3H2O = 2Al(OH)4

5) 2Na[Al(OH)4] + 3Na2SO4 = Al2(SO4)3 + 8NaOH

2Na + 2Al(OH)4 + 6Na + 3SO4 = 2Al + 3SO4 + 8Na + 8OH

2Al(OH)4 = 2Al + 8OH

6) Al2(SO4)3 + 6NaOH = 2Al(OH)3 + 3Na2SO4

2Al + 3SO4 + 6Na + 6OH = 2Al(OH)3 + 6Na + 3SO4

2Al + 6OH = 2Al(OH)3

7) Al(OH)3 +3HCl = 3H2O + AlCl3

Al(OH)3 + 3H + 3Cl = 3H2O + Al + 3Cl

Al(OH)3 + 3H = 3H2O + Al

8) AlCl3 + 4NaOH (избыток) = 3NaCl + Na[Al(OH)4]

Al + 3Cl + 4Na + 4OH = 3Na + 3Cl + Na + Al(OH)4

Al + 4OH = Al(OH)4