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GuDViN [60]
2 years ago
10

in an aerosol spray can with a volume of 456 ml contains is 3.18g of propane gas as a propellant If the can is at 23° C and 50 A

TM what volume will be the propane occupy at STP
Chemistry
1 answer:
Mekhanik [1.2K]2 years ago
8 0

Answer:    Answer:

21.03L

Explanation:

V1 = 456mL = 0.456L

T1 = 23°C = (23 + 273.15)K = 296.15K

P1 = 50atm

V2 = ?

T2 = 273.15K

P2 = 1.0atm

Note : P2 and T2 are at STP which are 1.0atm and 273.15K

To find V2, we have to use the combined gas equation,

(P1 × V1) / T1 = (P2 × V2) / T2

P1 × V1 × T2 = P2 × V2 × T1

V2 = (P1 × V1 × T2) / (P2 × T1)

V2 = (50 × 0.456 × 273.15) / (1.0 × 296.15)

V2 = 6227.82 / 296.15

V2 = 21.029L

Final volume of the gas is 21.03L

Explanation:

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What mass, in grams, of CO2 and H2O<br> is formed from 2.55 mol of propane?
oksian1 [2.3K]

Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

  • C₃H₈: 1 mole
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

Being the molar mass of each compound:

  • C₃H₈: 44 g/mole
  • O₂: 16 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

  • C₃H₈: 1 mole* 44 g/mole= 44 grams
  • O₂: 5 moles* 16 g/mole= 80 grams
  • CO₂: 3 moles* 44 g/mole= 132 grams
  • H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

  • If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}

mass of CO₂= 336.6 grams

  • If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}

mass of H₂O= 183.6 grams

<u><em>336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.</em></u>

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Answer:

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