Cinética
Considere a reação abaixo:
N2O5(g) → 2NO2(g) + 12 O2(g)
1. Inicialmente a concentração de N2O5 que esta dissolvido em tetracloreto
de carbono é 4,2 mol/L.
a) Qual a concentração inicial de NO2 e O2?
b) Após 1 h, ainda há N2O5 no recipiente e sua concentração é igual a
3,24mol/L. Calcule a velocidade média de decomposição da substância
no intervalo de 0 a 1 h.
c) Qual a concentração de N2O5, NO2 e O2 após 1 hora.
d) Após 2 h, a concentração de N2O5 e 2,40 mol/L. Calcule a velocidade
média de decomposição dessa substância no intervalo de 1 a 2 h.
e) Qual a concentração de N2O5, NO2 e O2 após 2 horas.
f) Após 3 h, a concentração de N2O5 é de 1,77. Calcule a velocidade média
de decomposição da substância no intervalo de 2 a 3 h.
g) Qual a concentração de N2O5, NO2 e O2 após 3 horas.
N205(9)
2NO2(g) + 12 O2(9)
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Ответ:
7.30g O₂ are in excess and 21.3g of SO₂ should be produced.
Explanation:
The reaction of Zinc sulfide (ZnS) with oxygen (O₂) produce ZnO and SO₂ as follows:
2ZnS + 3O₂ → 2ZnO + 2SO₂
Using this reaction we can find the amount of SO₂ produced converting the ZnS and O₂ to moles and finding limiting and excess reactant as follows:
Moles ZnS -Molar mass: 97.47g/mol-:
32.5g * (1mol / 97.47g) = 0.333 moles ZnS
Moles O₂ -Molar mass: 32g/mol-:
23.3g O₂ * (1mol / 32g) = 0.728 moles O₂
For a complete reaction of the 0.333 moles of ZnS are required:
0.333 moles ZnS * (3mol O₂ / 2mol ZnS) = 0.500 moles O₂.
As there are 0.728 moles of O₂, limiting reactant is ZnS and excess reactant is O₂.
The moles and mass of O₂ in excess are:
0.728mol - 0.500mol = 0.228moles O₂ * (32g / mol) = 7.30g O₂ are in excess
The produced mass of SO₂ -Obtained from the moles of ZnS- is:
0.333 moles of ZnS * (2mol SO₂ / 2mol ZnS) = 0.333 moles SO₂
0.333 moles SO₂ * (64.066g / mol) =
21.3g of SO₂ should be produced