Answer:
(240 × 3 × 31.998)/(122.5 × 2) g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 122.5 31.998
2KClO₃ ⟶ 2KCl + 3O₂
Mass/g: 240
Mass of O₂ = 240 g KClO₃ × (1 mol KClO₃/122.5 g KClO₃) × (3 mol O₂/2 mol KClO₃) × (31.998 g O₂/1 mol O₂) = 94.0 g O₂
Mass of O₂= (240 × 3 × 31.998)/(2 × 122.5) = 94.0 g O₂
The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g
mass of CO₂ formed is 591.8 g
Carbon has a tendency to form covalent bonds.