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Reacting Volumes

Reacting Volumes. Reactions of Gases. When we consider reactions of gases it is more convenient to use volumes instead of masses It is important to state the temperature & pressure for gaseous reactions because volume changes if these conditions are altered. Standard Conditions.

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Reacting Volumes

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  1. Reacting Volumes

  2. Reactions of Gases • When we consider reactions of gases it is more convenient to use volumes instead of masses • It is important to state the temperature & pressure for gaseous reactions because volume changes if these conditions are altered.

  3. Standard Conditions • Standard Temperature = 273K • K= Kelvin • 273K = 0oC • Standard Pressure = 100kPa • kPa = kiloPascal • 1kPa = 1 atmosphere

  4. 1 mole of any gas occupies 22.4dm3 at standard temperature & pressure (stp) • stp = 273K & 100kPa • 22.4dm3 is know as the molar volume • 1 mole of O2 weighs 32g (Mr = 2x16) • 1 mole of O2 at stp has a volume of 22.4 dm3 • So 32g of O2 occupies a volume of 22.4 dm3

  5. 134.4 dm3 of hydrogen is reacted with excess nitrogen in the presence of an iron catalyst. Assuming the reaction goes to completion what volume of NH3 is made? 1. Write balanced equation 3H2 + N2→ 2NH3 3 mole → 2 mole 2. Calculate no. moles of Hydrogen 1 mol of any gas at stp = 22.4 dm3 no. moles H2 = 134.4 22.4 = 6 mol

  6. 3. Calc no. moles NH3 from equation 3H2 + N2→ 2NH3 3 mole → 2 mole 1 mole → 2/3 mole No. moles NH3 = 2/3 x 6 = 4 4. Calculate volume of NH3 Volume of NH3 = 4 x 22.4 = 89.6 dm3

  7. 1. Write balanced equation • How many g of Mg will react with excess H2SO4 to produce 896 cm3 H2 at stp? Mg + H2SO4→ MgSO4 + H2 1 mole → 1 mole 2. Convert units of volume to dm3 896 cm3 = 896/1000 dm3 = 0.896 dm3 1 mol of any gas at stp = 22.4 dm3 no. moles H2 = 0.896 22.4 = 0.04 mol

  8. 3. Calc no. moles Mg from equation Mg + H2SO4→ MgSO4 + H2 1 mole → 1 mole No. moles Mg = 0.04 4. Calculate mass of Mg = No. moles x Mr Mass of Mg = 0.04 x 24 = 0.96g

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