A sample of oxygen occupies 20.1 liters under a pressure of 1520 torr at 25.0o What volume would it occupy at 25.0oC if the pres
1 answer:
Answer:
The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L
Explanation:
Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant, so that the pressure of a gas in a closed container is inversely proportional to the volume of the container. That is, if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
or P * V = k
Considering an initial state 1 and a final state 2, it is true:
P1* V1= P2*V2
In this case:
- P1= 20.1 L
- V1= 1520 torr
- P2= 760 torr
- V2= ?
Replacing:
20.1 L* 1520 torr= 760 torr* V2
Solving:
V2= 40.2 L
<em><u>The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L</u></em>
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Balanced chemical equation for the reaction is:
2S (g) +
(g)+ 2
O (l) ⇒
Moles of formed is 5.75 moles.
Moles of oxygen used is 5.75 moles in the reaction.
Explanation:
Data given:
moles of S = 11.5 moles
moles of = ?
Moles of needed =?
balanced equation with states of matter =?
Balanced chemical reaction under STP condition is given as:
2S(g) +
(g) + 2
O (l) ⇒
From the balanced reaction 2 moles of sulphur dioxide reacted to form 1 mole of sulphuric acid:
so, from 11.5 moles of S, x moles of
is formed
2x = 11.5
x = 5.75 moles of sulphuric acid formed.
From the balanced reaction 1 mole of oxygen reacted to form 1 mole of sulphuric acid.
when 11.5 moles of Sulphur dioxide reacted then oxygen in the reaction is 5.75 moles.