Question

ma puteți ajuta la pb 6 și IV?

Answer 1

6*. n=m/M=4,38/146=0,03 moli s.o

m/M=V/Vm=>m=M*V/Vm=28*2,016/22,4=2,52g CO

28g CO...................12g C
2,52g CO................x=1,08g C

m/M=V/Vm=>m=44*3,36/22,4=6,6g CO2

44g CO2.......................12g C
6,6g CO2.......................x=1,8g C

18g H2O.....................2g H2
4,86g H2O..................x=0,54g H2

mO=4,38-1,08-1,8-0,54=0,96g O

mC total = 1,8+1,08=2,88g

n=m/M=2,88/12=0,24 moli C

n=m/M=0,54/2=0,27 moli H2

n=m/M=0,96/16=0,06 moli O

0,03 moli s.o...........0,24 moli C.....0,27 moli H2........0,06 moli O
1 mol s.o................x=8 atomi C......y=18 atomi H......z=2 atomi O

F.M -> C8H18O2

IV. dgaz=M/Mgaz=>M=dgaz*Mgaz=2,49*28,9=72g/mol

n=m/M=1,44/72=0,02 moli s.o

m/M=V/Vm=>m=44*1,792/22,4=3,52g CO2

44g CO2..................12g C
3,52g CO2...............x=0,96g C

18g H2O....................2gH2
1,44g H2O.................x=0,16g H2

mO=1,44-0,96-0,16=0,32g 

n=m/M=0,96/12=0,08 moli C

n=m/M=0,16/2=0,08 moli H2

n=m/M=0,32/16=0,02 moli O

0,02 moli s.o...............0,08 moli C......0,08 moli H2........0,02 moli O
1 mol s.o.....................x=4 atomi C........y=8 atomi H......z=1 atom O

F.M => C4H8O

Metoda II

mCO2=3,52g

%C=300a/11s=300*3,52/11*1,44=66,666

%H=100b/9s=1,44*100/9*1,44=11,111

%O=100-(%C+%H) = 100-66,666-11,111=22,22

nC=%C*M/AC*100=66,666*72/1200=4 atomi C

nH=%H*M/AH*100=11,111*72/100=8 atomi H

nO=%O*M/AO*100=22,22*72/1600=1 atom O

F.M => C4H8O