Answer:
2.873 g of CO₂
Explanation:
This problem will be solved in two steps.
Step 1: Calculating mass of Octane:
Data Given:
Volume = 1 L = 1000 cm³
Density = 0.703 g/cm³
Mass = ??
Formula Used:
Density = Mass ÷ Volume
Solving for Mass,
Mass = Density × Volume
Mass = 0.703 g/cm³ × 1000 cm³
Putting Values,
Mass = 703 g
Step 2: Calculating Mass of Oxygen:
Data:
Volume = V = 5.0 L
Temperature = T = 25 °C = 298.15 K
Pressure = P = 1.0 atm
Moles = n = ?
Assuming that the gas is acting as Ideal gas so, we will use Ideal gas equation i.e.
P V = n R T
Solving for n,
n = P V / RT
Putting values,
n = 1.0 atm × 5.0 L / 0.0821 atm.L.mol⁻¹.K⁻¹ × 298.15 K
n = 0.204 moles
As,
Moles = Mass / M.Mass
So,
Mass = Moles × M.Mass
Mass = 0.204 mol × 16 g/mol ∴ M.Mass of O₂ = 16g.mol⁻¹
Mass = 3.26 g
Step 3: Calculating mass of CO₂:
The balance chemical equation is follow,
2 C₈H₁₈ + 25 O2 = 16 CO₂ + 18 H₂O
According to equation
228.45 g (2 mol) of C₈H₁₈ reacts with = 799.97 g (16 mol) of O₂
So,
703 g of C₈H₁₈ will react with = X g of O₂
Solving for X,
X = 703 g × 799.97 g ÷ 228.45
X = 2461 g of O₂
While, we are only provided with 3.26 g of O₂. This means O₂ is the limiting reactant and will control the yield of the final product. Therefore,
According to balance equation,
799.97 g (16 mol) of O₂ produced = 704.152 g (16 mol) of CO₂
So,
3.26 g (0.204 mol) of O₂ will produce = X g of CO₂
Solving for X,
X = 3.26 g × 704.152 g ÷ 799.97 g
X = 2.873 g of CO₂
