Aplicaciones De La Cantidad De Sustancia En La Cuantificación De Procesos Químicos.

Bloque I:
Aplicaciones de la cantidad de sustancia en la cuantificación de procesos químicos.

Masa Molar

H2O2= 2+32=34 gr/mol

CO2= 12+32=44 gr/mol

VolumenMolar

100gr C4H10
C4H10= 48+10=58 gr/mol
#mol=100/58= 1.72 mol
Volumen= (1.72 mol) (22.4 mol/L)= 38.52 L

Composición Porcentual

C6H14 = 72+14=86 gr/mol
%C=72/86(100)= 83.72%
%H= 14/86(100)= 16.18%

C6H12O6= 72+12+96= 180
%C=78/180(100)= 40%
%H=12/180(100)=6.66%
%O=96/180(100)=53.3%

Formula Mínima

C= 85.6%H=14.4%
C= 85.6/12 H=14.4/1
C=7.13 H=14.4
C=7.13/7.13 H=14.4/7.13
C= 1 H=2.01
CH2

C=40% H=6.66%O=53.33%
C= 40/12 H=6.66/1 O=53.33/16
C=3.33/3.33 H=6.66/3.33 O=3.33/3.33
C=1 H=2 O=1
CH2O

FormulaMolecular

C=40% H=6.66% O=53.33% pesa 180gr
C= 40/12 H=6.66/1 O=53.33/16
C=3.33/3.33 H=6.66/3.33 O=3.33/3.33
C=1H=2 O=1
CH2O= 12+2+16=30
180/30=6 6(CH2O)
C6H12O6

C=75% H=25% y pesa 32 gr
C=75/12 H=25/1
C=6.25/6.25H=25/6.25
C=1 H= 4
CH4 = 12+4=16
32/16=2 2(CH4)
C2H8

Balanceo Tanteo

4Cr+ 3O2 2Cr2O3
4 /2/0=Cr=2/4/ 4
6/2=O=3/6

Fe2O3 + 3CO 2Fe+3CO2
2=Fe=0/2
3/1=C=1/3
6/5/4=O=2/4/6

Balanceo Redox

SO2+ Br2+ 2H2O 2HBr+ H2SO4

+4S S+6 pierde e- (oxida)
0Br Br-1 gana e- (reduce)

2=Br=2
1=S=14=H=4
4=O=4

2CrCl3+ 3MnO2+ 2H2O 3MnCl2+ 2H2CrO4

+3Cr Cr+6 pierde e- (oxida)
+4Mn Mn+2 gana e- (reduce)

2=Cr=2
3=Mn=3
6=Cl=6
4=H=4
8=O=8