Question

I need help filing in the blanks with some explanation

Answer 1

299.39 grams

1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

100 g C₃H₈ 1 mol C₃H₈ 3 mol CO₂ 44.01 g CO₂

44.1 g C₃H₈ 1 mol C₃H₈ 1 mol CO₂

299.39 g CO₂

Step-by-step explanation

Given:

Mass of propane (C₃H₈) = 100 grams

What to find:

The mass of carbon dioxide (CO₂) produced.

Solution:

The balanced chemical equation for the combustion of propane, C₃H₈ is

1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

The next step is to convert 100 grams of propane (C₃H₈) to moles using the mole formula and molar mass of propane (C₃H₈).

`\begin{gathered} Moles=\frac{Mass}{Molar\text{ }mass} \\ \\ Molar\text{ }mass\text{ }of\text{ }propane\text{ }(C₃H₈)=44.1\text{ }g\text{/}mol \\ \\ Moles=\frac{100\text{ }g}{44.1\text{ }g\text{/}mol}=2.2676\text{ }mol\text{ }C₃H₈ \end{gathered}`

The step that follows is to determine the mole of carbon dioxide produced using the mole ratio from the equation and the mole of propane (C₃H₈) above.

`\begin{gathered} 1\text{ }mol\text{ }C₃H₈=3\text{ }mol\text{ }CO₂ \\ \\ 2.2676\text{ }mol\text{ }C₃H₈=x\text{ }mol\text{ }CO₂ \\ \\ x=\frac{2.2676\text{ }mol\text{ }C₃H₈}{1\text{ }mol\text{ }C₃H₈}*3\text{ }mol\text{ }CO₂ \\ \\ x=6.8028\text{ }mol\text{ }CO₂ \end{gathered}`

The final step is to convert 6.8028 moles of CO₂ to mass in grams of CO₂.

`\begin{gathered} Moles=\frac{Mass}{Molar\text{ }mass} \\ \\ Mass=Moles* Molar\text{ }mass \\ \\ Molar\text{ }mass\text{ }of\text{ }carbon\text{ }dioxide\text{ }(CO₂)=44.01\text{ }g\text{/}mol \\ \\ Mass=6.8028\text{ }mol*44.01\text{ }g\text{/}mol \\ \\ Mass=299.39\text{ }grams \end{gathered}`

Hence, the mass of carbon dioxide (CO₂) produced is 299.39 grams.

In summary,

1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

100 g C₃H₈ 1 mol C₃H₈ 3 mol CO₂ 44.01 g CO₂

44.1 g C₃H₈ 1 mol C₃H₈ 1 mol CO₂

299.39 g CO₂