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In a balanced chemical equation, the number that is written in front of the element or compound, shows the mole ratio in which the reactants combine to form a product. If there are no numbers in front of the element symbol, this means the number is '1'.

e.g. N 2 + 3 H 2 2 N H 3

In this reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

Calculate the molar mass of H 2 SO 4 .

  1. Hydrogen = 1.008 g.mol - 1 ; Sulfur = 32.07 g.mol - 1 ; Oxygen = 16 g.mol - 1

  2. M ( H 2 S O 4 ) = ( 2 × 1 . 008 ) + ( 32 . 07 ) + ( 4 × 16 ) = 98 . 09 g . m o l - 1
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Calculate the number of moles there are in 1kg of MgCl 2 .

  1. n = m M
    1. Convert mass into grams
      m = 1 k g × 1000 = 1000 g
    2. Calculate the molar mass of MgCl 2 .
      M ( M g C l 2 ) = 24 . 31 + ( 2 × 35 . 45 ) = 95 . 21 g . m o l - 1
  2. n = 1000 95 . 21 = 10 . 5 m o l

    There are 10.5 moles of magnesium chloride in a 1 kg sample.

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Barium chloride and sulfuric acid react according to the following equation to produce barium sulphate and hydrochloric acid.

B a C l 2 + H 2 S O 4 B a S O 4 + 2 H C l

If you have 2 g of BaCl 2 ...

  1. What quantity (in g) of H 2 SO 4 will you need for the reaction so that all the barium chloride is used up?
  2. What mass of HCl is produced during the reaction?
  1. n = m M = 2 208 . 24 = 0 . 0096 m o l
  2. According to the balanced equation, 1 mole of BaCl 2 will react with 1 mole of H 2 SO 4 . Therefore, if 0.0096 moles of BaCl 2 react, then there must be the same number of moles of H 2 SO 4 that react because their mole ratio is 1:1.

  3. m = n × M = 0 . 0096 × 98 . 086 = 0 . 94 g

    (answer to 1)

  4. According to the balanced equation, 2 moles of HCl are produced for every 1 mole of the two reactants. Therefore the number of moles of HCl produced is (2 × 0.0096), which equals 0.0192 moles.

  5. m = n × M = 0 . 0192 × 35 . 73 = 0 . 69 g

    (answer to 2)

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Group work : understanding moles, molecules and avogadro's number

Divide into groups of three and spend about 20 minutes answering the following questions together:

  1. What are the units of the mole? Hint: Check the definition of the mole.
  2. You have a 56 g sample of iron sulfide (FeS)
    1. How many moles of FeS are there in the sample?
    2. How many molecules of FeS are there in the sample?
    3. What is the difference between a mole and a molecule?
  3. The exact size of Avogadro's number is sometimes difficult to imagine.
    1. Write down Avogadro's number without using scientific notation.
    2. How long would it take to count to Avogadro's number? You can assume that you can count two numbers in each second.

Khan academy video on the mole - 1

More advanced calculations

  1. Calculate the molar mass of the following chemical compounds:
    1. KOH
    2. FeCl 3
    3. Mg(OH) 2
  2. How many moles are present in:
    1. 10 g of Na 2 SO 4
    2. 34 g of Ca(OH) 2
    3. 2.45 x 10 23 molecules of CH 4 ?
  3. For a sample of 0.2 moles of potassium bromide (KBr), calculate...
    1. the number of moles of K + ions
    2. the number of moles of Br - ions
  4. You have a sample containing 3 moles of calcium chloride.
    1. What is the chemical formula of calcium chloride?
    2. How many calcium atoms are in the sample?
  5. Calculate the mass of:
    1. 3 moles of NH 4 OH
    2. 4.2 moles of Ca(NO 3 ) 2
  6. 96.2 g sulfur reacts with an unknown quantity of zinc according to the following equation: Z n + S Z n S
    1. What mass of zinc will you need for the reaction, if all the sulfur is to be used up?
    2. What mass of zinc sulfide will this reaction produce?
  7. Calcium chloride reacts with carbonic acid to produce calcium carbonate and hydrochloric acid according to the following equation: C a C l 2 + H 2 C O 3 C a C O 3 + 2 H C l If you want to produce 10 g of calcium carbonate through this chemical reaction, what quantity (in g) of calcium chloride will you need at the start of the reaction?

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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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