13. Reacting Masses

Chemical reactions can be represented by balanced chemical equations. Balanced chemical
equations do ...
Sometimes the question mentions that one of the reactants is in excess. For example,
question 1 below says that “7.3g of h...
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13 Reacting Masses

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Chemical Calculations Worksheet

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13 Reacting Masses

  1. 1. 13. Reacting Masses Chemical reactions can be represented by balanced chemical equations. Balanced chemical equations do much more than tell us what reacts and what is formed. The big numbers that you see in front of some formulae are called stoichiometric coefficients. Mg + 2HCl → MgCl2 + H2 Stochiometric coefficients tell us the ratio of moles that react. In the example above, they tell us how much magnesium is needed to react completely with hydrochloric acid. When we read out the balanced equation above, we don’t just say ‘magnesium plus hydrochloric acid goes to magnesium chloride plus hydrogen’. Instead, we should actually say ‘one mole of magnesium plus two moles of hydrochloric acid goes to one mole of magnesium chloride plus one mole of hydrogen’. Calculating Reacting Masses There are three steps to these calculations: 1. Work out the number of moles of reactant (moles = mass / RAM). 2. Use the balanced chemical equation to work out the number of mass moles of product. 3. Work out the mass of product using mass = moles x RMM. moles RMM Example One Calculate the mass of magnesium chloride formed when 20g of magnesium reacts with hydrochloric acid: Mg + 2HCl → MgCl2 + H2 moles of Mg = mass / RAM = 20 / 24 = 0.833 mol moles of MgCl2 = 0.833 mol because the balanced chemical equation shows a 1:1 ratio mass of MgCl2 = 0.833 x (24 + 71) = 79.2g If you need to calculate the mass of reacant needed to make a given quantity of product, the method is almost identical: Example Two Calculate the mass of calcium carbonate needed to generate 11g of carbon dioxide according to the following balanced chemical equation: CaCO3 + H2SO4  CaSO4 + H2O + CO2 moles of CO2 = mass / RAM = 11 / (12 + 32) = 0.25 mol moles of CaCO3 = 0.25 mol because the balanced chemical equation shows a 1:1 ratio mass of CaCO3 = 0.25 x (40 + 12 + 48) = 25g
  2. 2. Sometimes the question mentions that one of the reactants is in excess. For example, question 1 below says that “7.3g of hydrogen chloride reacts with excess ammonium hydroxide “. This just means that all 7.3g of hydrogen chloride reacts and none is left over. 1) Calculate the mass of ammonium chloride, NH4Cl, formed when 7.3g of hydrogen chloride reacts with excess ammonium hydroxide: NH4OH + HCl → NH4Cl + H2O 2) Calcium carbide, CaC2, reacts with water to form acetylene, C2H2 : CaC2 + 2H2O → C2H2 + Ca(OH)2 Calculate the mass of acetylene formed when 80g of calcium carbide reacts with excess water. 3) Calculate the mass of sodium oxide, Na2O, formed when 1.2g of oxygen gas reacts with excess sodium: 4Na + O2 → 2Na2O 4) What mass of iron is formed when one kilogram (1 000 g) of iron ore, Fe2O3, reacts with excess carbon monoxide? Fe2O3 + 3CO → 2Fe + 3CO2 5) Calculate the mass of nitric acid required to react completely with 90g of copper: 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O 6) A student needs to prepare 80g of magnesium oxide, MgO, by combustion of magnesium metal in oxygen: Mg + H2O → MgO + H2 What mass of magnesium is required? 7) What mass of copper oxide needs to be reduced to obtain 8g of copper? CuO + H2 → Cu + H2O 8) Calculate the mass of sodium hydroxide required to precipitate out 36g of iron(II) hydroxide: FeSO4 + 2NaOH → Fe(OH)2 + Na2SO4 9) What mass of water is required to react with aluminium carbide, Al4C3, in order to form 9.6g of methane gas? Answers Al4C3 + 12H2O → 4Al(OH)3 + 3CH4 1. 10.7g 2. 32.5g 3. 4.65g 10) Calculate the mass of phosphoric acid required to generate 30g of 4. 700g hydrogen gas: 5. 236.25g 6Li + 2H3PO4 → 2Li3PO4 + 3H2 6. 48g 7. 10g 8. 32g 9. 43.2g 10. 196g

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