Question

Butane C₄H₁₀ is a common fuel found in cigarette lighters. When butane burns, it combines with oxygen to produce carbon dioxide and water. A student burned a 5.00-gram sample of butane. Show all calculations in solving the following problems.

1. Write the balanced equation for the combustion of butane.

2. Find the number of moles of butane in 5.00 grams.

3. Find the number of moles of oxygen that will react with 5.00 g of butane.

4. Find the number of grams of oxygen that will react with butane.

5. Find the number of moles of carbon dioxide produced.

6. Find the number of grams of carbon dioxide produces.

7. Find the number of moles of water produced.

8. Find the number of grams of water prduced.

9. Find the total mass of the reactants and compare it to the total mass of the products.

Answer 1

2C₄H₁₀ + 13O₂  ------------> 8CO₂  + 10H₂O

molar mass of butane = 58 g/mole

moles of 5 g of butane = 5/58 = 0.0862

Now, moles of O₂ required (13/2)*moles of butane = 0.5603 moles

Molar mass of O₂ = 32 g/mole

thus, mass of O₂ required = 0.5603*32 = 17.931 g

Moles of CO₂ produced = 4*moles of butane = 4*0.0862 = 0.3488

molar mass of CO₂ = 44 g/mole

thus, mass of CO₂ produced = 0.3488*44 = 15.1712 g

Moles of H₂O produced = 5*moles of butane = 0.431

Molar mass of H₂O= 18 g/mole

Thus,mass of H₂O produced = 0.431*18 = 7.758 g

Mass of reactants = mass of butane + mass of O₂ = 5 + 17.931 = 22.931 g

Mass of products = mass of CO₂ + Mass of H₂O = 15.1712 + 7.758 = 22.93 g

So, mass f products is equal Mass of reactants