Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 10, Problem 98AP
Interpretation Introduction

Interpretation:

The total volume and the partial pressures of the gases formed from the decomposition of nitroglycerin are to be calculated.

Concept Introduction:

The expression foran ideal gas equation is

PV=nRT.

Here, P is pressure, V is volume, n is moles, R is a gas constant, and T is temperature.

The expression to calculate mole fraction (χi) is

χi=nintotal.

Here, ni is the moles of component and ntotal is the total number of moles.

The relation between partial pressure (Pi) and total pressure (Ptotal) is:

χi=PiPtotal.

The number of moles is given by the expression, which is as:

n=given mass(m)molar mass(M).

The conversion factor is (29 mol product4 mol nitroglycerine).

So the moles can be calculated by the expression as:

n=given mass(m)molar mass(M)×(29 mol product4 mol nitroglycerine).

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

T(in K)=T(in °C)+273.15.

Expert Solution & Answer
Check Mark

Answer to Problem 98AP

Solution: V=1.7×102 L.

PCO2=0.49 atm.

PH2O=0.41 atm.

PN2=0.25 atm.

PO2=0.041 atm.

Explanation of Solution

Given information: 4C3H5(NO3)3(s)12CO2(g)+10H2O(g)+6N2(g)+O2(g).

Pressure is 1.2 atm, temperature is 25oC.

Mass of nitroglycerine is 2.6×102 g.

The mass of 1 mole of nitroglycerine is equal to its molar mass, that is 227.09 g.

1 mol nitroglycerine=227.09 g[molar mass of nitroglycerine=227.09 g/mol ].

Invert the conversion factor, whichbecomes (1 mol nitroglycerine227.09 g nitroglycerine).

According to the given reaction, the mole ratio of nitroglycerine and the products is

4 mol nitroglycerine=29 mol product.

Therefore, the conversion factor becomes (29 mol product4 mol nitroglycerine).

Multiply the conversion factors with the mass of nitroglycerine in order to calculate the moles of the products as

The moles can be calculated by the expression as

n=given mass(m)molar mass(M)×(29 mol product4 mol nitroglycerine)

Mol of products=(mol nitroglycerine227.09 g nitroglycerine)×(29 mol product4 mol nitroglycerine)×2.6×102 g nitroglycerine=8.3 mol.

Temperature is in Celsius. Convert it into Kelvin as:

T=(25+273) K=298 K.

The expression foran ideal gas equation is

PV=nRT.

Rearrange the above expression for V:

V=nRTP.

Substitute 8.3 mol for n, 0.0821 L atm/mol-K for R, 298 K for T, and 1.2 atm for P in the above expression as

V=8.3 mol×0.0821 L atm/mol-K×298 K1.2 atm=169.2 L.

As the minimum number of significant figures is 2, the final answer should also contain two significant figures, that is, 1.7×102 L.

The relation between partial pressure and total pressure is

Pi=χiPT. …… (1)

The formula to calculate the mole fraction is

χi=nintotal. …… (2)

Now, calculate the mole fractions of each product.

For CO2, substitute 12 mol for ni and 29 mol for ntotal in the above expression as

χi=nintotal

χCO2=12 mol29 mol=0.41.

For H2O, substitute 10 mol for ni and 29 mol for ntotal in equation (2) as

χi=nintotal

χH2O=10 mol29 mol=0.34.

For N2, substitute 6 mol for ni and 29 mol for ntotal in equation (2) as

χi=nintotal

χN2=mol29 mol=0.21.

For O2, substitute 1 mol for ni and 29 mol for ntotal in equation (2) as

χi=nintotal

χO2=mol29 mol=0.034.

For CO2, substitute 0.41 for χi and 1.2 atm for PT in equation (1) as

Pi=χiPT

PCO2=(0.41)(1.2 atm)=0.49 atm.

For H2O, substitute 0.34 for χi and 1.2 atm for PT in equation (1) as

Pi=χiPT

PH2O=(0.34)(1.2 atm)=0.41 atm.

For N2, substitute 0.21 for χi and 1.2 atm for PT in equation (1) as

Pi=χiPT

PN2=(0.21)(1.2 atm)=0.25 atm.

For O2, substitute 0.034 for χi and 1.2 atm for PT in equation (1) as

Pi=χiPT

PO2=(0.034)(1.2 atm)=0.041 atm.

Conclusion

The total volume of the gas is 1.7×102 L. The partial pressures of CO2, H2O, N2 and O2 are 0.49 atm ,0.41 atm ,0.25 atm and 0.041 atm, respectively.

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Chapter 10 Solutions

Chemistry

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