The reaction for the combustion of benzene is: 2c6h6 + 15o2 → 12 co2 +6 h2o. how moles of o2 are required to produce 130 g co2 in the presence of excess c6h6?

a.2.36
b.2.95
c.3.69
d.5.80

2 C₆H₆  +  15 O₂ → 12 CO₂  +   6 H₂O

if mass of CO₂ = 130g
since moles = mass ÷ molar mass

then moles of CO₂ = 130 g ÷ ((12 × 1) + (16 × 2))
                               =  2.9545 mol

Mole ratio of CO₂  :  O₂      is     12   :    15

∴ if mole of CO₂ = 2.9545 mol

then moles of O₂ = (2.9545 ÷ 12) × 15
                            =  3.6931 mol

∴ the answer is Option C

Because it was a dramatically big guy