Răspuns :
1.
2,3g 10-a md1
2Na + 2H2O --> 2NaOH + H2
2x23 2x18 2x40
=> 10-a = 2,3x2x18/2x23
= 10-a = 1,8 => 8,2 g apa ramasa
=> md = 2,3x2x40/2x23 = 4 g NaOH
=> ms1 = md1 + a = 4+8,2 = 12,2 g sol NaOH formata din reactie
ms2 = 20g , c2% = 10%
=> md2 = ms1xc1%/100
= 20x10/100 = 2 g NaOH
=> md.final = md1 + md2 = 4+2 = 6g
=> ms.final = ms1+ms2 = 12,2+20 = 32,2 g
=> c%.final= md.finalx100/ms.final
= 6x100/32,2 = 18,63%
2.
100%-5% = 95% puritate
19,2g Mg impur .......................... 100%
m.p. .............................................. 95%
= 18,24g Mg pur
18,24g m
2Mg + O2 --> 2MgO
2x24 2x40
=> m = 18,24x2x40/2x24 = 30,4 g MgO
30,4g m.final
MgO + H2O --> Mg(OH)2
40 58
=> m.final = 30,4x58/40 = 44,08 g Mg(OH)2