Question

a) What mass of carbon dioxide is produced from the complete combustion of 3.90×10⁻³ g of methane? Express your answer with the appropriate units.

b) What mass of water is produced from the complete combustion of 3.90×10⁻³ g of methane? Express your answer with the appropriate units.

c) What mass of oxygen is needed for the complete combustion of 3.90×10⁻³ g of methane? Express your answer with the appropriate units.

d) The fuel used in many disposable lighters is liquid butane, C4H10. How many carbon atoms are in 3.00 g of butane? Express your answer numerically in atoms.

Answer 1

a) What mass of carbon dioxide is produced from the complete combustion of 3.90 × 10⁻³ g of methane? Express your answer with the appropriate units.

CH₄ + 2O₂ ---> CO₂ + 2H₂O

Molar mass of CH₄ = 12 + 4 x 1 = 16 g / mole
Molar mass of CO₂ = 12 + 2 * 16 = 44 g / mole

16 g of CH₄ produces 44 g of CO₂.
3.90 x 10^-3 g of CH₄ produces = ( 44 x 3.90 x 10^-3) / 16
= 1.07 x 10^-2 g of CO₂

b) What mass of water is produced from the complete combustion of 3.90×10⁻³ g of methane? Express your answer with the appropriate units.

CH₄ + 2O₂ ---> CO₂ + 2H₂O

Molar mass of CH₄ = 12 + 4 x 1 = 16 g / mole
Molar mass of H₂O is = 2 x 1 + 16 = 18 g / mole

16 g of CH₄ produces = 2 x 18 = 36 g of H₂O

3.90 x 10^-3 g of CH₄ produces = ( 3.90 x 10^-3 x 36) / 16
= 8.78 x 10^-3 g of H₂O

c) What mass of oxygen is needed for the complete combustion of 3.90×10⁻³ g of methane? Express your answer with the appropriate units.​​​​​​​

CH₄ + 2O₂ ---> CO₂ + 2H₂O

Molar mass of CH₄ = 12 + 4 x 1 = 16 g / mole
Molar mass of O₂ is = 2 x 16 = 32 g / mole

16 g of CH₄ reacts with 2 x 32 g = 64 g of O₂
3.90 x 10^-3 g of CH₄ reacts = ( 64 x 3.90 x 10^-3) / 16
= 1.56 x 10^-2 g of O₂

d) The fuel used in many disposable lighters is liquid butane, C₄H₁₀. How many carbon atoms are in 3.00 g of butane? Express your answer numerically in atoms.​​​​​​​

1 mole of butane has 4 mole of carbon atom.

58 grams of C₄H₁₀ contains = 4 x 6.023 x 10²³ atoms of carbon

3.0 grams of C₄H₁₀ contains = (3 x 4 x 6.023 x 10²³) / 58 atoms of carbon

= 1.25 x 10²³ atoms of carbon