The total pressure of a mixture of hydrogen, oxygen, and methane is 96.4 kPa. The partial pressures of hydrogen and oxygen are 13.5
kPa and 29.3 kPa respectively. What is the partial pressure of methane?

A. 139.2 kPa
B. 53.6 kPa
C. 82.9 kPa
D. 42.8 kPa

Answers

Answer 1

Answer:

53.7kPa

Explanation:

Given parameters:

Total pressure of the mixture  = 96.4kPa

partial pressure of hydrogen  = 13.4kPa

partial pressure of oxygen  = 29.3kPa

Unknown:

partial pressure of methane  = ?

Solution:

The total pressure of mixture is equal to the sum of their particles pressures;

 

   Partial pressure of methane  = 96.4kPa - (13.4kPa + 29.3kPa)

  Partial pressure of methane = 53.7kPa

Answer 2

Answer: 53.6

Explanation: I just took the test and got it right.


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Answers

Your answer and workings are all mentioned in the picture below.

Taking into account the reaction stoichiometry, 56.42 grams of N₂ are formed when 4 moles of NH₃ react.

Reaction stoichiometry

In first place, the balanced reaction is:

4 NH₃ + 3 O₂  → 2 N₂+ 6 H₂O

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

NH₃: 4 molesO₂: 3 molesN₂: 2 molesH₂O: 6 moles

The molar mass of the compounds is:

NH₃: 17 g/moleO₂: 32 g/moleN₂: 28 g/moleH₂O: 18 g/mole

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

NH₃: 4 moles ×17 g/mole= 68 gramsO₂: 3 moles ×32 g/mole= 96 gramsN₂: 2 moles ×28 g/mole= 56 gramsH₂O: 6 moles ×18 g/mole= 108 grams

Mass of N₂ formed

The following rule of three can be applied: if by reaction stoichiometry 4 moles of NH₃ form 56 grams of N₂, 4.03 moles of NH₃ form how much mass of N₂?

[tex]mass of N_{2} =\frac{4.03 moles NH_{3} x56 grams of N_{2} }{4 moles NH_{3}}[/tex]

mass of N₂= 56.42 grams

Then, 56.42 grams of N₂ are formed when 4 moles of NH₃ react.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

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