Question

Raportul de masa si compozitia procentuala pentru:CO2,SO3,HNO3,H2O

Answer 1

1)raportul de masa:CO2=C:O=C:2×O=12:2×16=12:32=3:8
compoziția procentuala :1)miu CO2=Ac+2×Ao=12+2×16=12+32=44
2)44gCO2......12gC.....32gO
100gCO2.......x............y
x=12×100/44=27,27
y=32×100/44=72,72
2)So3
raport de masa =S:O =32:3×16=32:48=8:12
compoziția procentuala
1)miu So3=As+3×Ao=32+3×16 =32+48=80
2)80gSO3......32gS....48gO
100gSO3........x......y
x=32×100/80=40
y=48×100/80=60
3)HNO3
raport de masa H:N:O=1:14:3×16=1:14:48
compoziția procentuala
1)miu HNo3=Ah+An+3×Ao=1+14+3×16=1+14+48=15+48=63
2)63gHNO3....1gH....14gN..48gO
100g HNO3...x......y......z
x=100/63=1,58
y=14×100/63=22,22
z=48×100/63=76,19
4)raport de masa H2O
H:O=2:16=1:8
compoziția procentuala H2O
1)miu H2O=2×Ah+Ai=2×1+16=18
2)18gH2O.......2gH ....16g O
100g H2O....x......y
x=2×100/18=11,11
y=16×100/18=88,88

Answer 2

CO2  raport de masa  AC:2AO=12:2.16=12:32=3:8         SE SIMPLIF.CU 4
MCO2=12+2.16=44
44gCO2--------12gC-------32gO                    Compozitia proc.prin masa moleculara
100gCO2--------X-------------Y
X=100.12:44=27,27 la suta C
Y=100.32:44=72,72 la suta O

SO3       r.m      32:3.16=32:48=4:6=2:3
2+3=5
5gSO3---------2gS--------3gO
100gSO3--------a-----------b                  Comp.proc. prin raport de masa
a=100.2:5=40 la suta S
b=100.3:5=60 la suta O

HNO3         r.m.   1:14:3.16=1:14:48
1+14+48=63
63gHNO3------1gH--------14gN-------48gO
100g-------------a--------------b------------c
a=100:63=1,58 la suta H
b=100.14:63=22,22 la suta N
c=100.48:63=76,19 la suta O

H2O      r.a. 2:16=1:8
1+8=9
9g H2O-------1gH----------8gO
100g------------a-------------b
a=100:9=11,11 la suta H
b=100.8:9=88,88 la suta O