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svp [43]
3 years ago
9

A sample of gas initially has a volume of 1.42 L at 293 K and 1.30 atm. What volume will the sample have if the pressure changes

to 1.50 atm while the temperature is increased to 322 K?
Chemistry
1 answer:
34kurt3 years ago
8 0

1.35 litres is the volume will the sample have if the pressure changes to 1.50 atm while the temperature is increased to 322 K.

Explanation:

Data given:

initial volume in the sample pf gas V1 = 1.42 L

initial temperature of the sample of gas T1 = 293 K

Initial pressure of the gas P1 = 1.30 atm

final pressure of the sample of gas  P2= 1.50 atm

final temperature of the sample of gas T2 = 322 K

Final volume V2 = ?

following formula is used:

\frac{P1V1}{T1} = \frac{P2V2}{T2}

V2 = \frac{P1V1T2}{P2T1}

putting the values in the equation:

V2 = \frac{322 X 1.42 X 1.30}{1.50 X 293}

   = 1.35 Litre is the volume

When pressure and temperature of the gas is changed the volume becomes 1.35 litres.

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Using the following equation, 2C2H6 +7O2 --&gt;4CO2 +6H2O, if 2.5g C2H6 react with 170g of O2, how many grams of water will be p
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The mass of water (H₂O) that would be produced is 4.5 g

<h3>Stoichiometry </h3>

From the question, we are to determine the mass of water that would be produced.

From the given balanced chemical equation

2C₂H₆ +7O₂ → 4CO₂ +6H₂O

This means

2 moles of C₂H₆ reacts with 7 moles of O₂ to produce 4 moles of CO₂ and 6 moles of H₂O

Now, we will determine the number of moles of each reactant present

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Mass = 2.5 g

Molar mass = 30.07 g

Using the formula,

Number\ of\ moles = \frac{Mass}{Molar\ mass}

Number of moles of C₂H₆ present = \frac{2.5}{30.07}

Number of moles of C₂H₆ present = 0.08314 mole

  • For Oxygen (O₂)

Mass = 170g

Molar mass = 31.999 g/mol

Number of moles of O₂ present = \frac{170}{31.999}

Number of moles of O₂ present = 5.3127 moles

Since

2 moles of C₂H₆ reacts with 7 moles of O₂

Then,

0.08314 mole of C₂H₆ will react with \frac{7 \times 0.08314 }{2}

 \frac{7 \times 0.08314 }{2} = 0.58198 mole

Therefore,

0.08314 mole of C₂H₆ reacts with 0.58198 mole of O₂ to produce 3 × 0.08314 moles of H₂O

3 × 0.08314 = 0.24942 mole

Thus, the number of moles of water (H₂O) produced is 0.24942 mole

Now, for the mass of water that would be produced,

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of water = 18.015 g/mol

Then,

Mass of water that would be produced = 0.24942 × 18.015

Mass of water that would be produced = 4.4933 g

Mass of water that would be produced ≅ 4.5 g

Hence, the mass of water (H₂O) that would be produced is 4.5 g

Learn more on Stoichiometry here: brainly.com/question/14271082

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