Answer:
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
Answer:
Explanation:
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.
