Question

1. S-au obtinut 50 g solutie de potasa caustrica, de concentratie 8%.
Ce masa de potasa caustrica si ce masa de apa s-au folosit?
2. Calculati concentratia solutiei obtinuta prin dizolvarea a 6 g de zahar, in 14 ml de apa.
3. a) Care este masa de CO2 care contine cantitatea de oxigen ca si 4 moli de CaO?
b) Cate molecule de apa vor contine aceeasi cantitate de hidrogen ca si 3,4 g de NH3?

Answer 1

1. Cp=md/ms*100=>md=Cp*ms/100=8*50/100=4g KOH

ms=md+mH2O=>mH2O=ms-md=50-4=46g

2. roH2O=1g/cm3

ro=m/V=>m=ro*V=1*14=14g H2O

ms=md+mH2O=6+14=20g

md=6g

Cp=md/ms*100=6/20*100=30%

3.a)  1mol CaO............................16gO
4 moli CaO..............................x=64g O

1mol CO2...............................2*16=32g O2
x=2moli CO2....................................64g O2

mCO2=2*44=88g CO2

MCO2=44g/mol

b) MNH3=AN+3AH=14+3*1=17g/mol

n=m/M=3,4/17=0,2 moli NH3

mH=0,2*3=0,6g H

MH2O=2AH+AO=2*1+16=18g/mol

In 18g H2O se gasesc........................2gH
x=5,4g H2O....................................0,6g H

n=m/M=5,4/18=0,3 moli H2O

1mol H2O...........................6,022*10^23 molecule
0,3 moli..............................x=1,8066*10^23 molecule