A propane (C3H8) gas heater is used at a party. The propane tank was weighed before and after the event and it was calculated to have lost 150.0g of propane during the event. How many grams of carbon dioxide were produced by the heater?

Answer :

Answer

448.902 grams CO2

Explanation

Given:

The mass of propane lost = 150.0 g

What to find:

The grams of carbon dioxide produced.

Step-by-step solution:

The first step is to write a balanced chemical equation for the reaction.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

The next step is to convert 150.0 g C3H8 to moles using the mole formula

[tex]Mole=\frac{Mass}{Molar\text{ }mass}[/tex]

The molar mass of C3H8 = 44.1 g/mol

[tex]Mole=\frac{150.0\text{ }g}{44.1\text{ }g\text{/}mol}=3.40\text{ }mol\text{ }C_3H_8[/tex]

Now, use the mole ratio in the equation and the mole of propane lost to calculate the mole of carbon dioxide produced.

From the equation;

1 mole of CH₈ produces 3 moles of CO

Therefore, 3.40 moles of CH₈ will produce

[tex]\frac{3.40\text{ }mol\text{ }C_3H_8}{1\text{ }mol\text{ }C_3H_8}\times3\text{ }mol\text{ }CO_2=10.2\text{ }mol\text{ }CO_2[/tex]

The final step is to convert 10.2 moles of CO2 produced to grams using the mole formula in step 2 above.

Molar mass of CO2 = 44.01 g/mol

[tex]\begin{gathered} Mole=\frac{Mass}{Molar\text{ }mass} \\ \\ \Rightarrow Mass=Mole\times Molar\text{ }mass=10.2mol\times \\ \\ Mass=10.2\text{ }mol\times44.01g\text{/}mol=448.902\text{ }grams \end{gathered}[/tex]

Therefore, the grams of carbon dioxide produced is 448.902 grams

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