CHEM 212:CHEMISTSRY V 2
CHEM 212:CHEMISTSRY V 2
8th Edition
ISBN: 9781260304503
Author: SILBERBERG
Publisher: MCGRAW-HILL CUSTOM PUBLISHING
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Chapter 5.4, Problem 5.13BFP
Interpretation Introduction

Interpretation:

The volume of gaseous iodine pentafluoride measured 105°C and 0.935atm prepared by the reaction of 4.16g of solid iodine with 2.48L of gaseous fluorine at 18°C and 0.974atm is to be calculated.

Concept introduction:

Ideal gas law equation is as follows:

PV=nRT (1)

Here,

P is the pressure of the gas.

V is the volume occupied by gas.

n is the number of moles of gas.

T is the temperature in Kelvin.

R is the universal gas constant.

Rearrange equation (1) for n as follows:

n=PVRT (2)

Conversion factor to convert mL to L is as follows:

1mL=103L

The conversion factor to convert °C to Kelvin is as follows:

T(K)=T(°C)+273.15 (3)

When the number of each atom in the reactant side is equal toa number of each atom in the product side then the reaction is known as a balanced chemical reaction. The reactant present in the lower quantity of moles is the limiting reagent.

Expert Solution & Answer
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Answer to Problem 5.13BFP

The volume of gaseous iodine pentafluoride measured 105°C and 0.935atm prepared by the reaction of 4.16g of solid iodine with 2.48L of gaseous fluorine at 18°C and 0.974atm is 1.09L.

Explanation of Solution

The value of the universal gas constant (R) is 0.0821atmlitre/molK.

The pressure of fluorine gas (PF2) is 0.974atm.

 The volume (V) of the flask containing fluorine gas is 2.48L.

The weight of iodine gas is 4.16g.

The molecular weight of I2 is 253.8g/mol.

The value of the temperature of fluorine gas is 18°C.

The value of the temperature of IF5 is 105°C.

Substitute 18°C for CHEM 212:CHEMISTSRY V 2, Chapter 5.4, Problem 5.13BFP , additional homework tip  1 in equation (3) to calculate the temperature in Kelvin.

T(K)=(18°C+273.15)=291.15K

Substitute 105°C for CHEM 212:CHEMISTSRY V 2, Chapter 5.4, Problem 5.13BFP , additional homework tip  2 in equation (3) to calculate the temperature in Kelvin.

T(K)=(105°C+273.15)=378.15K

The balanced chemical equation for the given reaction is as follows:

I2(s)+5F2(g)2IF5(g)

Substitute 0.0821atmlitre/molK for R, 0.974atm for P, 2.48L for V, 291.15K for T in equation (2) to calculate the number of moles of fluorine gas.

nF2=(0.974atm)(2.48L)(0.0821atmlitremolK)(291.15K)=0.10105mol

According to the balanced chemical equation, the stoichiometric ratio between IF5 and F2 is 2:5.

The formula to calculate the number of moles of IF5 from F2 is as follows:

nIF5=(nF2)(2molIF55molF2) (4)

Substitute 0.10105mol for nF2 in equation (4).

nIF5=(0.10105mol)(2molIF55molF2)=0.040421mol

According to the balanced chemical equation, the stoichiometric ratio between IF5 and I2 is 2:1.

The formula to calculate the number of moles of IF5 from I2 is as follows:

nIF5=(mI2MI2)(2molIF51molI2) (5)

Substitute 4.16g for mI2 and 253.8g/mol for MI2 in equation (5).

nIF5=(4.16g253.8g/mol)(2molIF51molI2)=0.032782 mol

The amount of IF5 produced from I2 is less than the amount of IF5 produced from F2, therefore, I2 acts as a limiting reagent and amount of IF5 will be produced in accordance to the mole of I2.

Rearrange equation (2) for the volume of IF5 as follows:

VIF5=nRTP (6)

Substitute 0.032782 mol for n, 0.0821atmlitre/molK for  R, 378.15K for T and 0.935atm for P in equation (6).

VIF5=(0.0327mol)(0.0821atmlitremolK)(378.15K)(0.935atm)=1.0885L=1.09L

Conclusion

The volume of gaseous iodine pentafluoride measured 105°C and 0.935atm prepared by the reaction of 4.16g of solid iodine with 2.48L of gaseous fluorine at 18°C and 0.974atm is 1.09L.

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Chapter 5 Solutions

CHEM 212:CHEMISTSRY V 2

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