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CHEM- Gas Volume Calculations

Chemistry, Gas Volume Calculations, Molar Volume- IBDP | DSE | GCE | IAL | AP Chemistry

· chemistry,question analysis,Gas Volume,Molar Volume,Calculations

**Before moving onto analyzing some examples, here are few reminders.📝**

🧐**The volume occupied by one mole of a gas is called the molar volume which is 24 dm ^{3} / 24000 cm^{3}**

🧐**Number of moles of gas = Volume of gas / Molar Volume of gas**

*** Under the same temperature and pressure **❗️ ❕

**Example 1**👇

**What is the number of moles of methane in 1.44 dm ^{3} of the gas measured at room temperature and pressure? **

**(Molar Volume of gas at room temperature and pressure= 24 dm ^{3}mol^{-1})**

Answer: No of moles of methane= Volume of methane/Molar volume of gas

**Therefore, 1.44 dm ^{3} / 24 dm^{3}mol^{-1} = 0.0600 mol**

**Yup, keep in mind that for some questions all you need to do is substitute the correct values into the equation.**

***Watch out for the units too **❗️ ❕

**Example 2**👇

**What is the volume, measured at room temperature and pressure, occupied by 13.2 g of carbon dioxide?**

**(Relative atomic masses: C=12, O=16; Molar Volume of gas at room temperature and pressure= 24 dm ^{3}mol^{-1} )**

**Step 1:** **Find the moles of CO _{2}**

🧐**Moles= Mass / Molar Mass**

**Thus, moles of CO _{2} = 13.2 g / (12+16+16) = 0.300 mol**

💪**Final Step: ****Rearrange the formula to find the volume of gas**

🧐 **Volume of CO _{ 2} = Moles of CO_{2} × **

**Volume of CO _{ 2}= 0.300 mol × 24 dm^{3}mol^{-1} = 7.20 dm^{3}**

**Example 3**👇

**Butane burns in oxygen completely to produce carbon dioxide and water according to the following equation:**

**2C _{4}H_{10}(g) + 13O_{2}(g) ⇒ 8CO_{2}(g) + 10H_{2}O(l)**

**What volume of oxygen is needed for the complete combustion of butane measured at room temperature and pressure?**

(**Molar Volume of gas at room temperature and pressure= 24 dm ^{3}mol^{-1}= 24000 cm^{3} mol^{-1})**

**Step 1: Find the moles of C _{4}H_{10} using the formula**

**Moles of C _{4}H_{10}= Volume of C_{4}H_{10}/ Molar Volume of gas **

**Hence, 480 cm ^{3}/ 24000 cm^{3}mol^{-1}=0.02 mol**

**Step 2: Use mole ratio to find the moles of O _{2}**

🧐**From the equation, 2 moles of C _{4}H_{10} require 13 moles of O_{2}**

**Moles of O _{2} = 0.02 × (13/2) = 0.130 mol**

💪**Last Step: Rearrange the formula to find the volume of **O_{2}

🧐 **Volume of O _{2} = Moles of O_{2}**

**Volume of O _{2} = 0.130 mol × 24 dm^{3}mol^{-1} = 3.12 dm^{3}**

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