You are conducting a reaction in a 2.30 liter cylinder at 25.00C. How many grams of water (H2O, MM = 18.02 g/mol) will be produced if you combust methane gas (CH4, MM = 16.05 g/mol) that has a partial pressure of 2.20 atms with oxygen gas (O2, MM = 32.00 g/mol) that has a partial pressure of 3.30 atms in the 2.30 liter cylinder at 25.00C?

5.60g of water

Explanation:

The combustion of methane (CH₄), is:

CH₄ (g) + 2 O₂ (g) → CO₂ + 2 H₂O (g)

Using ideal gas law, PV = nRT. moles of CH₄ and O₂ are, respectively:

CH₄: [2.20atm * 2.30L] / [0.082 atmL/molK * 298K) = 0.207 moles

O₂: [3.30atm * 2.30L] / [0.082 atmL/molK * 298K) = 0.311 moles

As 1 mol of methane reacts with 2 moles of oxygen, the complete reaction of 0.311 moles of oxygen requires 0.311 / 2 = 0.155 moles of methane.

As you have 0.207 moles of methane, all oxygen will react. As 2 moles of oxygen produce 2 moles of water, produced moles of water are 0.311 moles H₂O. In grams:

0.311 moles H₂O * (18.02g / 1mol) = 5.60g of water

Equation:

CH4 + 2O2 - -> CO2 + 2H2O

Given:

Volume, V = 2.3 l

Temperature, T = 25 °C

Partial pressure of methane, Pm = 2.2 atm

Partial pressure of oxygen, Po = 3.3 atm

Note: R = 0.08206 (L. atm) / (mol. K).

Using ideal gas equation,

PmV = nmRT

nm = (2.2 * 2.3) / (298 * 0.08206)

= 0.2069 moles

Using ideal gas equation,

PoV = noRT

nm = (3.3 * 2.3) / (298 * 0.08206)

= 0.3104 moles

Finding the limiting reagent,

By stoichiometry, 1 mole of CH4 reacted with 2 moles of O2. Therefore, moles of O2 = 0.2069 mole of CH4/1 mole of CH4 * 2 moles of O2

= 0.4138 moles of O2 (> 0.3104 moles)

Oxygen is the limiting reagent.

Since 2 moles of O2 combusted to form 2 moles of water. Therefore, number of moles of H2O = 0.3104 moles

Molar mass of H2O = (2 * 1) + 16

= 18 g/mol

Mass = number of moles * molar mass

= 0.3104 * 18

= 5.59 g of H2O was produced.