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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

Respuesta :

JoshEast
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

Answer: 3.69 moles

Explanation: The molar mass of CO2 is 12 +16 +16 = 44 g/mole. The number of moles CO2 in 130 g = 130 g / (44 g/mole) = 2.95 moles CO2. The mole ratio of O2 consumed to CO2 produced is given in the equation as 15:12, or 5:4. So multiplication of moles CO2 by the mole ratio will give you the number of moles of O2 consumed: 2.95 moles CO2 X (5/4) = 3.69 moles O2 consumed.