Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 5, Problem 5.110P

(a)

Interpretation Introduction

Interpretation:

The grams of nickel that can be converted to the carbonyl with 3.55 m3 at 100.7 kPa is to be calculated.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

PV=constant

Here, P is the pressure and V is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

V α T

Here, T is the temperature and V is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

α n

Here, n is the mole of the gas and V is the volume.

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

(a)

Expert Solution
Check Mark

Answer to Problem 5.110P

The grams of nickel that can be converted to the carbonyl with 3.55 m3 at 100.7 kPa is 1.95×103 g.

Explanation of Solution

The equation for the reaction of Ni(aq) with CO(g) is as follows:

Ni(s) + 4CO(aq)Ni(CO)4(g)        (1)

The formula to convert °C to Kelvin is:

T(K)=T(°C)+273.15        (2)

Substitute 50 °C for T(°C) in equation (2).

T(K)=50+273.15=323.15 K

The expression to convert pressure from kPa into atm is as follows,

P= 700 torr(1 atm100.7 kPa)=0.993831729 atm

The expression to calculate the moles of the CO(g) is as follows,

PV=nRT        (3)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of CO(g) and R is the gas constant.

Rearrange the equation (3) to calculate n as follows,

n=PVRT        (4)

Substitute the value 0.993831729 atm for P, 323.15 K for T, 3.55 m3 for V and 0.0821 LatmmolK for R in the equation (4).

n=(0.993831729 atm)(3.55 m3)(1000 L1 m3)(0.0821 LatmmolK)(323.15 K)=1951.183 mol

From equation (1), four moles of the CO(g) reacts with one mole of Ni(s). Thus, the moles of Ni(s) is calculated from CO(g) as follows:

Moles of Ni(s)=[(Moles of CO(g))(1 mol Ni(s)4 mol CO(g))]        (5)

Substitute the value 1951.183 mol for moles of CO(g) in the equation (5).

Moles of Ni(s)=[(1951.183 mol)(1 mol Ni(s)4 mol CO(g))]=487.79 mol

The expression to calculate the mass of Ni(s) is as follows:

Moles of Ni=Mass of NiMolar mass of Ni        (6)

Rearrange the equation (6) to calculate the mass of Ni(s) is as follows:

Mass of Ni=(Moles of Ni)(Molar mass of Ni)        (7)

Substitute the value 487.79 mol for moles of Ni(s) and 58.69 g/mol for molar mass of Ni(s) in the equation (7).

Mass of Ni=(487.79 mol)(58.69 g/mol)=1951.183 g1.95×103 g

Conclusion

The grams of nickel that can be converted to the carbonyl with 3.55 m3 at 100.7 kPa is 1.95×103 g.

(b)

Interpretation Introduction

Interpretation:

The mass of nickel obtained per cubic meter of the carbonyl at 155 °C and 21 atm is to be calculated.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

PV=constant

Here, P is the pressure and V is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

V α T

Here, T is the temperature and V is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

α n

Here, n is the mole of the gas and V is the volume.

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

(b)

Expert Solution
Check Mark

Answer to Problem 5.110P

The mass of nickel obtained per cubic meter of the carbonyl at 155 °C and 21 atm is 3.5×104 g.

Explanation of Solution

The formula to convert °C to Kelvin is:

T(K)=T(°C)+273.15        (8)

Substitute 155 °C for T(°C) in equation (8).

T(K)=155 °C+273.15=428.15 K

The expression to calculate the moles of the Ni(CO)4 is as follows,

PV=nRT        (9)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of Ni(CO)4 and R is the gas constant.

Rearrange the equation (9) to calculate n as follows,

n=PVRT        (10)

Substitute the value 21 atm for P, 428.15 K for T, 1 m3 for V and 0.0821 LatmmolK for R in the equation (10).

n=(21 atm)(1 m3)(1000 L1 m3)(0.0821 LatmmolK)(428.15 K)=597.42059 mol

From the equation (1), one mole of the Ni(s) will be formed one mole of Ni(CO)4. Thus, the moles of Ni(s) is calculated from Ni(CO)4 as follows:

Moles of Ni(s)=[(Moles of Ni(CO)4)(1 mol Ni(s)2 mol Ni(CO)4)]        (11)

Substitute the value 597.42059 mol for moles of CO(g) in the equation (11).

Moles of Ni(s)=[(597.42059 mol)(1 mol Ni(s)1 mol Ni(CO)4(g))]=597.42059 mol

The expression to calculate the mass of Ni(s) is as follows:

Moles of Ni=Mass of NiMolar mass of Ni        (12)

Rearrange the equation (12) to calculate the mass of Ni(s) is as follows:

Mass of Ni=(Moles of Ni)(Molar mass of Ni)        (13)

Substitute the value 597.42059 mol for moles of Ni(s) and 58.69 g/mol in the equation (13).

Mass of Ni=(597.42059 mol)(58.69 g/mol)=3.50626×104 g3.5×104 g

Conclusion

The mass of nickel obtained per cubic meter of the carbonyl at 155 °C and 21 atm is 3.5×104 g.

(c)

Interpretation Introduction

Interpretation:

The volume of CO(g) at 35 °C and 769 torr is to be calculated.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

PV=constant

Here, P is the pressure and V is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

V α T

Here, T is the temperature and V is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

α n

Here, n is the mole of the gas and V is the volume.

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

(c)

Expert Solution
Check Mark

Answer to Problem 5.110P

The volume of CO(g) formed at 35 °C and 769 torr is 63 m3.

Explanation of Solution

The expression to calculate the mass of Ni(s) is as follows:

Moles of Ni=Mass of NiMolar mass of Ni        (14)

Substitute 3.50626×104 g for the mass of Ni(s) and 58.69 g/mol in the equation (14).

Moles of Ni=(3.50626×104 g)(58.69 g/mol)=597.420 mol

From the equation (1), four moles of the CO(g) reacts with one mole of Ni(s). Thus, the moles of CO(g) is calculated from Ni(s) as follows:

Moles of CO(g)=[(Moles of CO(g))(1 mol CO(g)4 mol Ni(s))]        (15)

Substitute the value 597.420 mol for moles of CO(g) in the equation (15).

Moles of CO(g)=[(597.420 mol)(4 mol CO(g)1 mol Ni(s))]=2389.68238 mol

Substitute 35 °C for T°C in equation (8).

T(K)=35 °C+273.15=308.15 K

The expression to calculate the pressure of CO(g) is as follows,

Ptotal=Pwater vapour+PCO        (16)

Rearrange the equation (16) to calculate the PCO.

PCO=PtotalPwater vapour        (17)

Substitute 42.2 torr for Pwater vapour and 769 torr for Ptotal in the equation (17).

PCO=769 torr42.2 torr=726.8 torr

The expression to convert torr into atm is as follows,

P= 726.8 torr(1 atm760 torr)=0.956315789 atm

The expression to calculate the volume of CO(g) is as follows,

PV=nRT        (18)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of CO(g) and R is the gas constant.

Rearrange the equation (18) to calculate V as follows,

V=nRTP        (19)

Substitute the value 0.993831729 atm for P, 308.15 K for T, 2389.68238 mol for n and 0.0821 LatmmolK for R in the equation (4).

V=(2389.68238 mol)(0.0821 LatmmolK)(308.15 K)(0.956315789 atm)=(63218.4995 L)(1 m31000 L)=63.2184995 m363 m3

Conclusion

The volume of CO(g) at 35 °C and 769 torr is 63 m3.

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Chapter 5 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

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