Question

Care ma poate ajuta si pe mine? O proba de sodiu contine 16,56325•10 la puterea 24 protoni, si se adauga intr-un vas ce contine 200g solutie NaOH de concentratie 30%. Calculati concentratia solutiei finale obtinute.

Answer 1

-calculez moli protoni

niu=N/N,A= 16,56325X10²⁴/6,023X10²³mol= 27,5mol protoni

calculez moli atomi Na, stiind ca 1 atom are 11 protoni

niu,Na=2,5 mol Na

calculez masa de sodiu adaugata,

m= niuxM= 2,5molx23g/mol= 5,75gNa; aceasta masa reactioneaza cu apa din solutia in care se adauga

-calculez masa de apa si masa de hidroxid ,din ecuatia chimica

2x23g  ....2x18g..........2x40g

2Na +  2H2O --------> 2NaOH  + H2

5,75g........x.................y

x= 4,5g H2O

y= 10gNaOH

DECI 4,5 g apa se scade din apa existenta in solutia data iar 10g NaOH se adauga la masa existenta in acea solutie

c=mdx100/ms----> md= 30x200/100g= 60gNaOH

m,apa =ms-md= 200-60=>140g

m,apa ramasa= 140-4,5g=> 135,5g

solutia finala are:  md= md,initial +md,format=60+10=> 70g

                              ms= md+ m,apa= 70g+ 135,5g= 205,5g

c,final= 70x100/205,5=...................calculeaza !!!!!