6 reacting mass calculations

• In this equation:

  N2 + 3 H2 -> 2 NH3

  Means that there is 1 mole of N2, 3 moles of H2 and 2 moles NH3. So if the number of moles in N2 increased by a multiple of 12 it would now mean that there are 12 moles of N2, 36 moles of H2 and 24 moles of NH3
• You can use balanced chemical equations to find out what mass of chemicals (or volume of gases) react or are produced in a chemical reaction. To do this calculate:

  1. Moles of /
  2. Moles of ?
  3. Mass of ?

  E.g what mass of iron is produced when 32.0kg of iron (III) oxide is heated with CO

  / ?

  Fe2O3(S) + 3CO(g) -> 2Fe(s) + 3CO2(g)

  Moles of Fe2O3 = mass (g) ÷ mass = 32000 ÷ 159.6 = 200.5 mol

  1 mole of Fe2O3 forms 2 Moles of Fe

  Mples of Fe = 2 × 200.5 = 401.0 mol

  Mass of Fe = Moles × Mr = 401.0 × 55.8 = 22,400g (3SF)
• Another example of renting mass calculations

  what mass Oxygen is needed to convert 102g of ammonia into nitrogen?

  / ?

  4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(g)

  Moles of NH3 = mass ÷ Mr =

  102 ÷ 17.0 = 6.00mol

  4 moles of Nh3 reacts with 3 moles of O2 therefore 1 mole of NH3 reacts with 3/4 mole od O2 therefore moles of O2 = 6.00 × 3/4 = 4.50 mol

  Therefore mass of O2 = moles × Mr = 4.50 × 32.0 = 144g (3 sig fig)
• Another example of reacting mass calculations

  When 5.00g of crystals of hydrated tin (II) chloride, SnCl2.xH2O are heated 4.20g of anhydrous tin (II) chloride are formed. Calculate the number of molecules of water of crystallisation are in SnCl2.xH20 I.e the value of x

  SnCl2.xH20 -> SnCl2 + xH20

  Moles of SnCl2 = mass ÷ Mr = 4.20 ÷ 189.7 = 0.02214 moles therefore the moles of SnCl2.xH20 = 0.02214

  Therefore the Mr of SnCl2.xH2O = mass ÷ moles = 5.00 ÷ 0.02214 = 225.8 therefore the Mr of xH20 = 225.8 - 189.7 = 36.1 therefore

  x = 36.1 ÷ 18.0 = 2

  X is a whole number and so the final answer is given as an integar
• Usually we have more than we need of one of the reactants and o it doesn't all react - it is in excess
• Sometimes in calculations we need to work out if one of the reactants I in excess. The reactant that us not in excess is sometimes called the limiting reagent
• An example of a limiting rwleagent question

  Propane reacts with oxygen as shown C3H8 + 5O2 -> 3CO2 + 4H2O

  How many moles of products are formed when 1 mole of C3H8 is mixed with 8 moles of O2

  C3H8 + 5O2 -> 3CO2 + 4H2O

  1 8

  1 react 5 react 3 made 4 made

  1-1=0. 8-5=3. 0+3=3. 0+4=4

  O2 in excess

  C3H8 is the limiting reagent
• In calculations you will be asked to work out masses but you will need to convert to moles to find out which is the limiting reagent in order to work out the required answer

  E.g in the manufacture of titanium what mass of titanium can theoretically be formed when 1.00kg of titanium chloride reacts with 0.100kg of magnesium

  TiCl4 + 2Mg -> Ti + 2MgCl2

  Moles of TiCl4 at the start 1000÷189.9 = 5.266mol

  Moles of Mg at the start 100÷24.3 = 4.115mol

  5.266 moles of TiCl4 needs 10.53 moles of Mg to react therefore TiCl4 is in excess and does not all react so Mg is the limiting reagent therefore 2.058 moles of TiCl4 reacts with 4.115 moles of Mg

  Change in moles for TiCl4 = -2.058

  Change in moles for Mg = -4.115

  Change in moles for Ti = 2.058

  Change in moles 2MgCl2 = 4.115

  Therefore mass of Ti = 2.058 × Mr = 2.058 × 47.9 = 98.6g
• When you make a new substance by a chemical reaction, you may not get all the expected amount of product example, if you reacted 4g of hydrogen with 32 g of oxygen, you may get less than 36 g of water Reasons include

  • the reaction may be reversible (both the forwards and backwards reaction can take place)
  • some of the product may be lost when it is separated from the reaction mixture
  • some of the reactants may react in other reactions.
• %yield = (mass of product obtained ÷ maximum theoretical mass of product ) × 100
• Percentage yield calculation example

  Tungsten is made from tungsten(VI) oxide

  WO3 + 3H2 -> W + 3H2O

  A ) calculate the maximum theoretical mass of tungsten that can be made from 0.500 tonne of tungsten(VI) Oxide

  Moles of WO3 = mass (g) ÷ Mr = 500000 ÷ 231.8 = 2157mol therefore moles of W = 2157 mol therefore the mass of W = moles × Mr = 2157×183.8 = 396000g (3 sig fig)

  B) in the reaction only 350000g of tungsten was made. Calculate the percentage yield

  %yield = (350000 ÷ 396000) × 100 = 88.3% (3 sig fig)
• Atom economy is a measure of what proportion of the products of a reaction are the desired product and how much is waste. The higher the atom economy the less waste that is produced
• %atom economy = (mass of desired products as shown in equation ÷ total mass of product as shown in equation) × 100
• Example of an atom economy question

  Glucose -> ethanol + carbon dioxide

  C6H12O6(aq) -> 2C2H7OH(aq) + 2CO2

  Glucose = 180g

  Ethanol = 92g

  Carbon dioxide = 88g

  Atom economy = (92÷180) × 100 = 51%

  Only 92g of the 180g of products is ethanol. This means that 51% of the mass of the products is ethanol while the other 49% is waste