Question

What mass of oxygen is needed to burn 54.0 g of butane

Answer 1

The mass of oxygen needed to burn 54.0g of butane is 193.24 g

Step-by-step explanation

Given:

Mass of butane = 54.0 g

What to find:

The mass of oxygen needed to burn 54.0g of butane.

Step-by-step solution:

Step 1: Write the balanced chemical equation for the reaction.

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Step 2: Convert 54.0 g of butane to moles

Molar mass of C₄H₁₀ = 58.12 g/mol

`Moles=\frac{Mass}{Molar\text{ }mass}=\frac{54.0\text{ }g}{58.12\text{ }g\text{/}mol}=0.9291\text{ }mol`

Step 3: Use the mole ratio in step one and the mole of C₄H₁₀ in step 2 to calculate the mole of oxygen.

`\begin{gathered} 13\text{ }mol\text{ }O_2=2\text{ }mol\text{ }C_4H_(10) \\ \\ x=0.9291\text{ }mol\text{ }C_4H_(10) \\ \\ x=\frac{0.9291\text{ }mol\text{ }C_4H_(10)}{2\text{ }mol\text{ }C_4H_(10)}*13\text{ }mol\text{ }O_2 \\ \\ x=6.03915\text{ }mol\text{ }O_2 \end{gathered}`

Step 4: Convert the moles of O₂ in step 3 to mass in grams.

Molar mass of O₂ = 31.998 g/mol

`\begin{gathered} Mass=Moles* Molar\text{ }mass \\ \\ Mass=6.03915mol*31.998g\text{/}mol \\ \\ Mass=193.24\text{ }g\text{ }of\text{ }O_2 \end{gathered}`

Hence, the mass of oxygen needed to burn 54.0g of butane is 193.24 g