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Stoichiometric calculations

Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It is also the numerical relationship between reactants and products. In representing chemical change showed how to write balanced chemical equations. By knowing the ratios of substances in a reaction, it is possible to use stoichiometry to calculate the amount of either reactants or products that are involved in the reaction. The examples shown below will make this concept clearer.

What volume of oxygen at S.T.P. is needed for the complete combustion of 2 dm 3 of propane ( C 3 H 8 )? (Hint: CO 2 and H 2 O are the products in this reaction (and in all combustion reactions))

  1. C 3 H 8 ( g ) + 5 O 2 ( g ) 3 C O 2 ( g ) + 4 H 2 O ( g )

  2. From the balanced equation, the ratio of oxygen to propane in the reactants is 5:1.

  3. 1 volume of propane needs 5 volumes of oxygen, therefore 2 dm 3 of propane will need 10 dm 3 of oxygen for the reaction to proceed to completion.

What mass of iron (II) sulphide is formed when 5,6 g of iron is completely reacted with sulphur?

  1. Fe ( s ) + S ( s ) FeS ( s )

  2. n = m M = 5 , 6 55 , 85 = 0 , 1 mol
  3. From the equation 1 mole of Fe gives 1 mole of FeS . Therefore, 0,1 moles of iron in the reactants will give 0,1 moles of iron sulphide in the product.

  4. m = n × M = 0 , 1 × 87 , 911 = 8 , 79 g

    The mass of iron (II) sulphide that is produced during this reaction is 8,79 g .

When we are given a known mass of a reactant and are asked to work out how much product is formed, we are working out the theoretical yield of the reaction. In the laboratory chemists never get this amount of product. In each step of a reaction a small amount of product and reactants is 'lost' either because a reactant did not completely react or some of the product was left behind in the original container. Think about this. When you make your lunch or supper, you might be a bit hungry, so you eat some of the food that you are preparing. So instead of getting the full amount of food out (theoretical yield) that you started preparing, you lose some along the way.

Sulphuric acid ( H 2 SO 4 ) reacts with ammonia ( NH 3 ) to produce the fertiliser ammonium sulphate (( NH 4 ) 2 SO 4 ) according to the following equation:

H 2 SO 4 ( aq ) + 2 NH 3 ( g ) ( NH 4 ) 2 SO 4 ( aq )

What is the maximum mass of ammonium sulphate that can be obtained from 2,0 kg of sulphuric acid?

  1. n ( H 2 SO 4 ) = m M = 2 000 g 98 , 078 g · mols - 1 = 20 , 39 mols
  2. From the balanced equation, the mole ratio of H 2 SO 4 in the reactants to ( NH 4 ) 2 SO 4 in the product is 1:1. Therefore, 20,39 mols of H 2 SO 4 of ( NH 4 ) 2 SO 4 .

    The maximum mass of ammonium sulphate that can be produced is calculated as follows:

    m = n × M = 20 , 41 mol × 132 g · mol - 1 = 2694 g

    The maximum amount of ammonium sulphate that can be produced is 2,694 kg .

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Source:  OpenStax, Chemistry grade 10 [caps]. OpenStax CNX. Jun 13, 2011 Download for free at http://cnx.org/content/col11303/1.4
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