The volume of oxygen required to burn 12.00 L ethane is calculated as follows
find the moles of C2H6 used
At STP 1 mole is always = 22.4 L, what about 12.00 L
= ( 12.00L x 1 moles) 22.4 L = 0.536 moles
write the reacting equation
2C2H6+ 7O2 = 4CO2 + 6H2O
by use of mole ratio between C2H6 :O2 which is 2:7 the moles of O2
= 0.536 x7/2= 1.876 moles
again at STP 1mole = 22.4 L what about 1.876 moles
= 22.4 L x 1.876 moles/ 1 mole = 42.02 L
The mass of oxygen reacted/required in this reaction is obtained as 48g.
<h3>What is stoichiometry?</h3>
The term stoichiometry has to do with mass- volume or mass - mole relationship which ultimately depends on the balanced reaction equation.
Now, we have the reaction; S + O2 ------>SO2
If 1 mole of sulfur dioxide contains 22.4 L
x moles of sulfur dioxide contains 33.6L
x = 1.5 moles of sulfur dioxide.
Since the reaction is 1:1, the number if moles of oxygen required/reacted is 1.5 moles.
Mass of oxygen required/reacted = 1.5 moles * 32 g/mol = 48g
Learn more anout stoichiometry: brainly.com/question/9743981
Answer:
does your son ever talk to strangers?
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>
Answer:
The correct answer is option D which is the decreasing order of conductivity is Mn, O, Ge.
Explanation:
You can easily answer this if you know the periodic trends. For the property of electrical conductivity, it decreases across a period and decreases also down a group. Thus, the most conductive element must be Mn, while the least conductive one is Ge. So, the answer is: -Mn, O, Ge
