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6.1 Stoichiometry and composition  (Page 4/5)

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Molarity and the concentration of solutions

  1. 5.95 g of potassium bromide was dissolved in 400 cm 3 of water. Calculate its molarity.
  2. 100 g of sodium chloride (NaCl) is dissolved in 450 cm 3 of water.
    1. How many moles of NaCl are present in solution?
    2. What is the volume of water (in dm 3 )?
    3. Calculate the concentration of the solution.
    4. What mass of sodium chloride would need to be added for the concentration to become 5.7 mol.dm - 3 ?
  3. What is the molarity of the solution formed by dissolving 80 g of sodium hydroxide (NaOH) in 500 cm 3 of water?
  4. What mass (g) of hydrogen chloride (HCl) is needed to make up 1000 cm 3 of a solution of concentration 1 mol.dm - 3 ?
  5. How many moles of H 2 SO 4 are there in 250 cm 3 of a 0.8M sulphuric acid solution? What mass of acid is in this solution?

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. The Grade 10 discussion on representing chemical changeshowed 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.

Khan academy video on stoichiometry - 1

What volume of oxygen at S.T.P. is needed for the complete combustion of 2dm 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.

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What mass of iron (II) sulfide is formed when 5.6 g of iron is completely reacted with sulfur?

  1. F e ( s ) + S ( s ) → F e S ( s )

  2. n = m M = 5 . 6 55 . 85 = 0 . 1 m o l

  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 sulfide in the product.

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

    The mass of iron (II) sulfide that is produced during this reaction is 8.79 g.

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A closer look at the previous worked example shows that 5.6 g of iron is needed to produce 8.79 g of iron (II) sulfide. The amount of sulfur that is needed in the reactants is 3.2 g. What would happen if the amount of sulfur in the reactants was increased to 6.4 g but the amount of iron was still 5.6 g? Would more FeS be produced? In fact, the amount of iron(II) sulfide produced remains the same. No matter how much sulfur is added to the system, the amount of iron (II) sulfide will not increase because there is not enough iron to react with the additional sulfur in the reactants to produce more FeS. When all the iron is used up the reaction stops. In this example, the iron is called the limiting reagent . Because there is more sulfur than can be used up in the reaction, it is called the excess reagent .
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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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