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Chapter 5: Problem 142

A 20.0 -L stainless steel container at \(25^{\circ} \mathrm{C}\) was charged with 2.00 atm of hydrogen gas and 3.00 atm of oxygen gas. A spark ignited the mixture, producing water. What is the pressure in the tank at $25^{\circ} \mathrm{C}$ ? If the exact same experiment were performed, but the temperature was \(125^{\circ} \mathrm{C}\) instead of \(25^{\circ} \mathrm{C},\) what would be the pressure in the tank?

Short Answer

Expert verified
After performing the calculations, the total pressures in the stainless steel container after the reaction at the given temperatures are approximately \(1.48\,\mathrm{atm}\) at \(25^{\circ}\mathrm{C}\) and \(2.18\,\mathrm{atm}\) at \(125^{\circ}\mathrm{C}\).

Step by step solution

01

Write down the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between hydrogen gas and oxygen gas to produce water is: \[ 2\mathrm{H}_{2(g)} + \mathrm{O}_{2(g)} \rightarrow 2\mathrm{H}_{2}\mathrm{O}_{(g)} \]
02

Calculate the number of moles of hydrogen and oxygen initially

Using the ideal gas law (in terms of moles), rearrange the formula to solve for n: \[ n = \frac{PV}{RT} \] The initial number of moles of hydrogen gas (n_H2) and oxygen gas (n_O2) can be calculated: \(n_{\mathrm{H}_2} = \frac{P_{\mathrm{H}_2}V}{RT}\), \(n_{\mathrm{O}_2} = \frac{P_{\mathrm{O}_2}V}{RT}\), where \(P_{\mathrm{H}_2} = 2.00\,\mathrm{atm}\) and \(P_{\mathrm{O}_2}= 3.00\,\mathrm{atm}\) are the pressures, V is the volume of the container (20 L), R is the gas constant (0.0821 L atm mol\(^{-1}\) K\(^{-1}\)), and T is the temperature in Kelvin: \(T_1 = 25^{\circ} \mathrm{C}+273.15 = 298.15\,\mathrm{K}\) and \(T_2 = 125^{\circ} \mathrm{C}+273.15 = 398.15\,\mathrm{K}\).
03

Find the limiting reactant

To determine the limiting reactant, we need to check the mole ratios of hydrogen and oxygen. From the balanced chemical equation, we have the mole ratio \(\frac{\mathrm{H}_2}{\mathrm{O}_2} = \frac{2}{1}\). Calculate the effective available moles of hydrogen (n_H2_eff) considering this ratio: \(n_{\mathrm{H}_2}^{\mathrm{eff}} = 2n_{\mathrm{O}_2}\). If \(n_{\mathrm{H}_2}^{\mathrm{eff}} < n_{\mathrm{H}_2}\), then oxygen will be the limiting reactant. Otherwise, hydrogen is the limiting reactant. We find the limiting reactant by comparing these values.
04

Calculate the moles of hydrogen and oxygen remaining and the pressure after the reaction

Based on the limiting reactant, calculate the moles of hydrogen and oxygen remaining after the reaction, which will be n_H2_remaining and n_O2_remaining. The total pressure (P_Total) after the reaction can then be calculated as the sum of the individual partial pressures: \[ P_{\mathrm{Total}} = P_{\mathrm{H}_2}^{\mathrm{remaining}} + P_{\mathrm{O}_2}^{\mathrm{remaining}}, \] where \(P_{\mathrm{H}_2}^{\mathrm{remaining}} = \frac{n_{\mathrm{H}_2}^{\mathrm{remaining}}RT}{V}\), \(P_{\mathrm{O}_2}^{\mathrm{remaining}} = \frac{n_{\mathrm{O}_2}^{\mathrm{remaining}}RT}{V}\). Calculate the total pressure for the two different temperatures provided in the exercise.
05

Write down your results

After following all the steps mentioned, you should have the pressures in the tank after the reaction, for the given temperatures (\(25^{\circ} \mathrm{C}\) and \(125^{\circ} \mathrm{C}\)). Make sure to state your answers in atm or any other appropriate pressure unit.

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