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Genrish500 [490]
1 year ago
13

A 5. 0 l flask containing o2 at 2. 00 atm is connected to a 3. 0 l flask containing h2 at 4. 00 atm and the gases are allowed to

mix. What is the mole fraction of h2? report your answer to two decimal places.
Chemistry
1 answer:
dsp731 year ago
3 0

The mole fraction of hydrogen is 0.55.

The volume of oxygen = 5 litres

The pressure of oxygen = 2 atm

The volume of hydrogen = 3 litres

The pressure of hydrogen = 4 atm

The universal gas constant is R.

R = 8.31 J mol—¹ K—¹

The ideal gas equation is,

PV = nR T

The number of moles of oxygen is,

PV = nRT

n =  \frac{ PV }{RT }

n =  \frac{2 \times 5}{8.31 \times 273}

n = 0.004 moles

The number of moles of oxygen is 0.004 moles.

The number of moles of hydrogen is,

PV = nRT

n =  \frac{ PV }{RT }

n =  \frac{4 \times 3}{8.31 \times 273}

n = 0.005 moles

The number of moles of hydrogen is 0.005 moles.

The mole fraction of hydrogen is,

Mole \:  fraction =  \frac{ Moles \:  of \:  solute }{Total \:  number  \: of  \:  moles  \: of  \: solute \:  and \:  solvent}

Mole \:  fraction  =  \frac{0.005}{0.009}

= 0.55

Therefore, the mole fraction of hydrogen is 0.55.

To know more about mole fraction, refer to the below link:

brainly.com/question/28344281

#SPJ4

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Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

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Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

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Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

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