Determinați raportul atomic raportul de masa Compoziție procentuala pt urmatoarele substanțe
a)SO2
b)Mg(OH)2.
C) KMnO4
2. Calculați cantitatea de 5 continuta in 40 g SO2 130gH2S. 5 moliH2SO4
Răspunsuri la întrebare
SO2
r.a.=1 : 2
r.m.=32 : 32= 1: 1
comp. proc.prin masa molara
M=32+32=64-------> 1mol=74g [ 64g/moli ]
64g SO2-------32g S--------32g O
100g --------------x----------------y
x=100.32 :64=50% S
y=50% O
Mg( OH ) 2
r.a.=1 : 2 : 2
r.m.=24 : 2.16 : 2
=24 : 32 :2 [se simpl. cu 2 ]
=12 : 16 : 1
M=24 +32 +2=58-------> 58g/moli
58g Mg(OH )2--------24g Mg--------32gO---------2g H
100g----------------------x---------------------y---------------z
x=100.24 :58=41,38 % Mg
y=100.32 : 58=55,17 % O
z=100.2 : 58=3,45 % H
KMnO4
r.a.=1 : 1 : 4
r.m.= 39 : 55 : 64
M=39 +55 +64=158------>158g/moli
158 g---------39gK--------55gMn--------64g O
100--------------x-----------------y----------------z
x=100.39 :158=24,69% K
y=100.55 :158=34,81%Mn
z=100.64 :158=40,51 %O
2.
40g SO2
g S
MSO2=32 +32=64------> 64g/moli
64g SO2---------32gS
40g ------------------x
x=40 .32 :64=20 g S
130g H2S
g S
M=2+32 =34-------> 34g/moli
34 g H2S---------32 gS
130 g-----------------x
x=130.32 :34=122,35 g S
5moli H2SO4
g,moli S
-se afla masa de H2SO4 din 5 moli
n=m:M [n=nr.moli m=masa subst. M=masa molara ]
m=n.M
m=5moli .98g/moli=490 g H2SO4
M=2+32+ 64=98------->98g/moli
98g H2SO4---------32 gS
490 g ------------------xg
x=490 .32 :98=160g S
n=160g : 32g/moli=5moli S