Answer:
This is a coal combustion process and we will assume
Inlet coal amount = 100kg
It means that there are
15kg of H2O, 2kg of Sulphur and 83kg of Carbon
Now to find the mole fraction of SO2(g) in the exhaust?
Molar mass of S = 32kg/kmol
Initial moles n of S = 2/32 = 0.0625kmols
Reaction: S + O₂ = SO₂
That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂
Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust
The mole fraction of SO2(g) in the exhaust=0.0625kmols
Explanation:
Answer:
3.33 tanques de O₂
Explanation:
Basados en la reacción:
2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)
<em>2 moles de acetileno reaccionan con 5 moles de oxígeno produciendo 4 moles de dióxido de carbono y 2 moles de agua</em>
<em />
La ley de Avogadro dice que el volumen de un gas bajo temperatura y presión constantes es proporcional a las moles de este gas. Así, como 2 moles de acetileno reaccionan con 5 moles de oxígeno, los litros de O₂ necesarios para quemar 9340L de acetileno son:
9340 L C₂H₂ × (5 moles O₂ / 2 moles C₂H₂) = <em>23350L de O₂</em>
Si un tanque contiene 7x10³ L de O₂ serán necesarios:
23350L O₂ ₓ (1 tanque / 7x10³L) =<em> 3.33 tanques de O₂</em>
The mass number of aluminium hydroxide is 78 thus, the number of moles in 0.745 g is:
no. of moles= mass/ RFM
= 0.745/78
=0.00955moles
Therefore the 0.00955 moles should be in the 35.18 ml
therefore 1000ml of the solution will have:
(0.00955ml×1000ml)/35.18
=0.2715moles
The solution will be 0.27M hydrochloric acid
Answer:
a) heat it from 23.0 to 78.3
q = (50.0 g) (55.3 °C) (2.46 J/g·°C) =
b) boil it at 78.3
(39.3 kJ/mol) (50.0 g / 46.0684 g/mol) =
c) sum up the answers from the two calculations above. Be sure to change the J from the first calc into kJ
Explanation:
