Question

1.Cate grame de apa sunt intr-un cristalizator,daca introducand 0,5 moli de atomi de sodiu in cristalizator se formeaza o solutie de concentratie 30%
2.Se arde o masa impura de 3 g aluminiu cu 1,68 L oxigen.Aflati puritatea aluminiului(impuritatile sunt inerte).
3.Peste 5,75 g de sodiu se adauga 94,5 g de apa.Determinati concentratia procentuala a solutiei
4.10 g solutie H2SO4 de concentratie 98% reactioneaza total cu 2,5 g magneziu impur.Sa se calculeze puritatea magneziului

Answer 1

1. MNa=23g/mol

n=md/M=>md=n*M=0,5*23=11,5g Na

C=md/ms*100=>ms=md*100/C=11,5*100/30=38,333g sol

ms=md+mH2O=>mH2O=ms-md=38,33-11,5=26,833g

2. 4Al + 3O2 => 2Al2O3

4*27g Al...........3*22,4 L O2
xg Al.................1,68 L O2

x=(4*27*1,68)/(3*22,4) = 2,7g Al (m pur)

p=mp/mimp*100=2,7/3*100=90℅

3. Na + H2O => NaOH + 1/2H2

23g Na....18g H2O....40g NaOH...0,5*2g H2
5,75g Na...x=4,5g....y=10g......z=0,25g

mdf = 5,75+10+0,25=16g

msf = 4,5+16+94,5=115g sol

Cf=mdf/msf*100=16/115*100=13,913℅

4. C=md/ms*100=>md=C*ms/100=98*10/100=9,8g H2SO4

n=m/M=9,8/98=0,1 moli H2SO4

24g.......1 mol
Mg + H2SO4 => MgSO4 + H2
x=2,4g.....0,1 moli

p=mp/mimp*100=2,4/2,5*100=96℅