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Chapter 9: Problem 77

A flask with a volume of \(10.0\) L contains \(0.400 \mathrm{~g}\) of hydrogen gas and \(3.20 \mathrm{~g}\) of oxygen gas. The mixture is ignited and the reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} $$ goes to completion. The mixture is cooled to \(27^{\circ} \mathrm{C}\). Assuming \(100 \%\) yield, (a) What physical state(s) of water is (are) present in the flask? (b) What is the final pressure in the flask? (c) What is the pressure in the flask if \(3.2 \mathrm{~g}\) of each gas is used?

Short Answer

Expert verified
Answer: When 0.400 g of hydrogen gas and 3.20 g of oxygen gas react, the final pressure in the flask is approximately 0.734 atm, and the physical state of water is gaseous. When 3.2 g of each gas is used, the pressure in the flask is approximately 7.73 atm, and the physical state of water remains gaseous.

Step by step solution

01

Find the limiting reactant

Convert grams of hydrogen gas and oxygen gas to moles: - Using molar mass: H2 = 2.02 g/mol, O2 = 32 g/mol Moles of H2 = (0.400 g) / (2.02 g/mol) = 0.198 moles Moles of O2 = (3.20 g) / (32 g/mol) = 0.100 moles Now, compare the mole ratios: H2 reaction ratio = 0.198 / 2 (from balanced equation) O2 reaction ratio = 0.100 / 1 (from balanced equation) Since O2 has a smaller ratio, it is the limiting reactant.
02

Calculate the theoretical yield of water

Using the balanced equation, 1 mole of O2 reacts with 2 moles of H2 to produce 2 moles of H2O. So, we can calculate the moles of H2O produced: Moles of H2O = 2 * moles of O2 = 2 * 0.100 = 0.200 moles
03

Determine the remaining amount of reactant

Since H2 is not completely consumed in the reaction, calculate the remaining moles: Moles of H2 remaining = moles of H2 - 2 * moles of O2 = 0.198 - 2 * 0.100 = 0.098 moles
04

Calculate the final pressure in the flask

Use the ideal gas law (PV = nRT) to find the pressure of the system. The total moles in the flask after the reaction are 0.098 moles of H2 remaining + 0.200 moles of H2O. Total moles = 0.098 + 0.200 = 0.298 moles Given volume of the flask is 10 L and after cooling the mixture temperature is 27°C = 27 + 273.15 = 300.15 K. R (ideal gas constant) = 0.0821 L atm / (K mol) Now, use PV = nRT to find the pressure (P): P = (nRT) / V = (0.298 mol * 0.0821 L atm / (K mol) * 300.15 K) / 10 L = 0.734 atm So, the final pressure in the flask is approximately 0.734 atm.
05

Determine the physical state(s) of water

At the given temperature (27°C), water is in the gaseous state. Thus, only gaseous water is present in the flask.
06

Calculate the pressure in the flask if 3.2 g of each gas is used

In this case, moles of H2 = (3.2 g) / (2.02 g/mol) = 1.58 moles Moles of O2 = (3.20 g) / (32 g/mol) = 0.100 moles Here, H2 is the limiting reactant. The theoretical yield of H2O: Moles of H2O = 2 * moles of H2 = 2 * 1.58 = 3.16 moles No O2 remains in the flask, and there are 3.16 moles of H2O present. Using the ideal gas law (PV = nRT), we can find the pressure: P = (nRT) / V = (3.16 mol * 0.0821 L atm / (K mol) * 300.15 K) / 10 L = 7.73 atm So, the pressure in the flask if 3.2 g of each gas is used is approximately 7.73 atm.

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Most popular questions from this chapter

19\. Argon gas has its triple point at \(-189.3^{\circ} \mathrm{C}\) and \(516 \mathrm{~mm} \mathrm{Hg}\). It has a critical point at \(-122^{\circ} \mathrm{C}\) and \(48 \mathrm{~atm}\). The density of the solid is \(1.65 \mathrm{~g} / \mathrm{cm}^{3}\) whereas that of the liquid is \(1.40 \mathrm{~g} / \mathrm{cm}^{3}\). Sketch the phase diagram for argon and use it to fill in the blanks below with the words "boils" "melts" "sublimes," or "condenses." (a) Solid argon at \(500 \mathrm{~mm} \mathrm{Hg} \) when the temperature is increased. (b) Solid argon at 2 atm increased. (c) Argon gas at \(-150^{\circ} \mathrm{C}\) when the pressure is increased. (d) Argon gas at \(-165^{\circ} \mathrm{C} \) when the pressure is increased.

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