Question

If 12 grams of butane (C4H10), a fuel in cigarette lighter, reacts with 12 grams of O2, how

many grams of the products will be formed?

 Balanced Equation: 2𝐶4𝐻10 + 13𝑂2 → 8𝐶𝑂2 + 10𝐻2𝑂

Answer 1

Molar mass of butane =58.12

12/58.12=0.21mol

Molar mass of O2= 31.998

12/31.998=0.38mol

Molar mass of CO2= 44.01

Mole ratio CO2:O2= 8:13

Mole ratio CO2: C4H10=8:2

= 4.615×44.01×0.59=119.83g

Molar mass of H2O=18.01528

Mole ratio H2O:O2 =10:13

Mole ratio H2O: C4H10=10:2

5.77×18.01528×0.59=61.33g