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Chapter 20: Problem 100

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) can be produced by thermal decomposition of ammonium nitrate: $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \stackrel{\mathrm{heat}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ What volume of \(\mathrm{N}_{2} \mathrm{O}(g)\) collected over water at a total pressure of 94.0 \(\mathrm{kPa}\) and \(22^{\circ} \mathrm{C}\) can be produced from thermal decomposition of 8.68 $\mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{3} ?\( The vapor pressure of water at \)22^{\circ} \mathrm{C}$ is 21 torr.

Short Answer

Expert verified
The volume of \(N_2O\) that can be collected over water after thermal decomposition of 8.68 g of \(NH_4NO_3\) at a total pressure of 94.0 kPa and 22°C is approximately 2.843 liters.

Step by step solution

01

Convert mass of \( NH_4NO_3 \) to moles

The molar mass of \( NH_4NO_3 \) is \( 14 + 1\cdot4 + 14 + 16\cdot3 = 80\, g/mol \). To convert the given mass (8.68 g) into moles, divide it by its molar mass: \[ \frac{8.68\,g}{80\, g/mol} = 0.1085\, moles \]
02

Mole ratio calculation

From the reaction, it can be seen that 1 mole of \( NH_4NO_3 \) will produce 1 mole of \( N_2O \). Hence the amount of \( N_2O \) produced is also 0.1085 moles.
03

Getting the pressure of \( N_2O \)

The total collected pressure is the sum of the partial pressure of the \( N_2O \) and the water vapor. Given that the vapor pressure of water at \( 22^{\circ}C \) is 21 torr, which is equivalent to 2.8 kPa, the pressure of \( N_2O \) can be found as follows: \[ 94.0\,kPa - 2.8\,kPa = 91.2\,kPa \]
04

Convert the temperature to Kelvin

To use the ideal gas law, the temperature must be in Kelvin. \( 22^{\circ}C \) is equivalent to \( 22 + 273.15 = 295.15\,K \)
05

Apply the ideal gas law to find the volume of \( N_2O \)

The ideal gas law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is number of moles, \( R \) is the universal gas constant, and \( T \) is temperature. We can rearrange the formula to solve for volume (\( V = \frac{nRT}{P} \)). Substituting the known values: \[ V = \frac{0.1085\,moles \cdot 8.314\,L\cdot kPa/(mol\cdot K) \cdot 295.15\,K}{91.2\,kPa} = 2.843\,L \] So, the volume of \( N_2O \) that can be produced from the decomposition of 8.68 g of \( NH_4NO_3 \) under the given conditions is approximately 2.843 liters.

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