Chapter 6 Gases - 6.7
- Volume and Moles (Avogadro’s Law)
- In Avogadro’s Law
- the volume of a gas is directly related to the number of moles (n) of gas.
- T and P are constant.
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Learning Check - If 0.75 mole helium gas occupies a volume of 1.5 L, what volume will 1.2 moles helium occupy at the same temperature and pressure?
- 1) 0.94 L
- 2) 1.8 L
- 3) 2.4 L
Solution - 3) 2.4 L
- STEP 1 Conditions 1 Conditions 2
- V1 = 1.5 L V2 = ???
- n1 = 0.75 mole He n2 = 1.2 moles He
- STEP 2 Solve for unknown V2
- V2 = V1 x n2
- n1
- STEP 3 Substitute values and solve for V2.
- V2 = 1.5 L x 1.2 moles He = 2.4 L
- 0.75 mole He
STP - The volumes of gases can be compared at STP, Standard Temperature and Pressure, when they have
- the same temperature.
- Standard temperature (T)
- 0°C or 273 K
- the same pressure.
- Standard pressure (P)
- 1 atm (760 mm Hg)
- At standard temperature and pressure (STP), 1 mole of a gas occupies a volume of 22.4 L, which is called its molar volume.
Molar Volume as a Conversion Factor - The molar volume at STP can be used to form conversion factors.
- 22.4 L and 1 mole
- 1 mole 22.4 L
Using Molar Volume - What is the volume occupied by 2.75 moles N2 gas
- at STP?
- The molar volume is used to convert moles to liters.
- 2.75 moles N2 x 22.4 L = 61.6 L
- 1 mole
Guide to Using Molar Volume Learning Check - A. What is the volume at STP of 4.00 g of CH4?
- 1) 5.60 L 2) 11.2 L 3) 44.8 L
- B. How many grams of He are present in 8.00 L of gas at STP?
- 1) 25.6 g 2) 0.357 g 3) 1.43 g
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Solution - A. 1) 5.60 L
- 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L
- 16.0 g CH4 1 mole CH4
- B. 3) 1.43 g
- 8.00 L x 1 mole He x 4.00 g He = 1.43 g He
- 22.4 L 1 mole He
- The volume or amount of a gas at STP in a chemical
- reaction can be calculated from
- STP conditions.
- mole factors from the balanced equation.
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STP and Gas Equations - What volume (L) of O2 gas is needed to completely
- react with 15.0 g of aluminum at STP?
- 4 Al(s) + 3 O2 (g) 2 Al2O3(s)
- Plan: g Al mole Al mole O2 L O2 (STP)
- 15.0 g Al x 1 mole Al x 3 moles O2 x 22.4 L (STP)
- 27.0 g Al 4 moles Al 1 mole O2
- = 9.33 L O2 at STP
Learning Check - What mass of Fe will react with 5.50 L O2 at STP?
- 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)
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Solution - 4Fe(s) + 3O2(g) 2Fe2O3(s)
- ? 5.50 L at STP
- 5.50 L O2 x 1 mole x 4 moles Fe x 55.9 g Fe = 18.3 g
- 22.4 L 3 moles O2 1 mole Fe
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