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How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume that there is excess Xe. Xe(g) +3F2(g)→XeF6(g) A) 8.25 × 1023 molecules XeF6 B) 1.21 × 1023 molecules XeF6 C) 1.37 × 1023 molecules XeF6 D) 7.29 × 1023 molecules XeF6 E) 2.75 × 1023 molecules XeF6

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Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number  (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

                                  = 2.75 x 10²³ molecules XeF₆.

To determine the number of molecules of XeF₆,

First, we will determine the number of moles of F₂ present in the 12.9L of F₂

From the ideal gas equation

PV = nRT

Where P is the pressure

V is the volume

n is the amount of substance ( number of moles)

R is ideal gas constant

T is the temperature

From the question,

P = 2.6 atm

V = 12.9 L

R = 0.082057 L atm mol⁝š K⁝š

T = 298 K

Putting these values into the equation

PV = nRT

2.6 × 12.9 = n × 0.082057 × 298

33.54 = n × 24.452986

∴ n = 33.54 á 24.452986

n = 1.3716116 moles

The number of moles of F₂ present is 1.3716116 moles

From the given equation of reaction

Xe(g) +3F₂(g)→XeF₆(g)

1 mole of Xe reacts with 3 moles of F₂ to produce 1 mole of XeF₆

∴ 1.3716116 moles of F₂ will give (1.3716116/3) moles of XeF₆

Hence, the number of moles of XeF₆ produced = 0.457204 moles

To determine the number of molecules formed,

From the formula

Number of molecules = number of moles × Avogadro's number

∴ Number of molecules of XeF₆ formed = 0.457204 × 6.02214 ×10²³

Number of molecules of XeF₆ formed = 2.7533 × 10²³ molecules

Number of molecules of XeF₆ formed ≅ 2.75 × 10²³ molecules

Hence, the number of molecules of XeF₆ that are formed from 12.9 L of F₂ (at 298 K and 2.6 atm) is 2.75 × 10²³ molecules.

The correct option is E) 2.75 × 1023 molecules XeF6

Learn more about calculating number of molecules here: https://brainly.com/question/24231329

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