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What volume would be occupied by two moles of propane gas, C3H8, at standard conditions?

User Sanae
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1 Answer

20 votes
20 votes

Answer:

44.8L

Step-by-step explanation:

In order to get the volume of the gas that would be occupied by two moles of propane gas, we need to first calculate the mass of two moles of propane gas.

Determine the required mass of propane


\begin{gathered} \text{Mole}=\frac{\text{Mass}}{\text{Molar mass}} \\ Mass\text{ of propane=mole}* molar\text{ mass} \\ \text{Mass of propane=2moles}*\lbrack3(12)+8(1)\rbrack \\ \text{Mass of propane}=2\cancel{\text{moles}}*\frac{44g}{\cancel{\text{mole}}} \\ \text{Mass of propane}=88g \end{gathered}

At standard temperature and pressure (STP)

44g of the gas will occupy 22.4L volume. Hence 88g of propane gas will occupy:


\begin{gathered} \text{volume of propane=}\frac{\cancel{88g}^2*22.4L}{\cancel{44g}} \\ \text{volume of propane}=2*22.4L \\ \text{volume of propane}=44.8L \end{gathered}

Therefore the volume that would be occupied by two moles of propane gas, C3H8, at standard conditions is 44.8L.

User Ofornes
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