1 Chemical Reactions and Equations

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1 1 Intext Questions On Page 6 Question 1. Why should a magnesium ribbon be cleaned before burning in air? Magnesium ribbon reacts with CO present in air to form a protective and inert layer of magnesium carbonate. This layer is unreactive and hence, needs to be cleaned before burning in air. Question. Write the balanced equation for the following chemical reactions. (i) Hydrogen chlorine hydrogen chloride (ii) Barium chloride + aluminium sulphate barium sulphate + aluminium chloride (iii) Sodium + water sodium hydroxide hydrogen (i) The symbol/formulae of hydrogen, chlorine and hydrogen cholride are H, Cl and HCl respectively. Thus, the simplest equation is H Cl HCl Reactant Product Now the balancing is done in following steps. Step I Count the number of atoms of each elements on the reactant and product side of the equation. LHS RHS H 1 Cl 1 Step II In order to equalate the number of H atoms, write in the equation before HCl. Now the equation becomes. Step III sides. H + Cl HCl Count further the number of atoms of each element on both At LHS At RHS No. of H atoms No. of Cl atoms No. of H atoms No. of Cl atoms Now the number of atoms of each element are equal on both the sides. Thus, H() g Cl HCl( g) is the balanced equation for the given reaction. (Where, g represents the gaseous state).

2 (ii) Formula of barium chloride, aluminium sulphate, barium sulphate and aluminium chloride is respectively BaCl,Al (SO 4),BaSO 4, AlCl. So, the unbalanced equation is BaCl Al (SO 4) BaSO4 AlCl on balancing the above equation [by the similar method as shown for (i)] we get BaCl Al (SO 4) BaSO 4+ AlCl (iii) Similary balanced equation for the give reaction is Na + HO NaOH + H Question. Write a balanced chemical equation with state symbols for the following reactions. (i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride. (ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to form sodium chloride solution and water. (i) In this reactants are barium chloride (BaCl ) and sodium sulphate (NaSO 4) and products are barium sulphate (BaSO 4) and sodium chloride (NaCl). So the balanced equation is BaCl( aq) NaSO4( aq) BaSO4 () s NaCl ( aq) Barium chloride Sodium sulphate Barium sulphate Sodium chloride (ii) In this reaction, sodium hydroxide (NaOH) and hydrogen chloride (HCl) are reactants and sodium chloride (NaCl) and water (HO) are the products so, the balanced equation for this reaction is NaOH ( aq) HCl ( aq) NaCl ( aq) HO () l Sodium hydroxide Hydrogen chloride Sodium chloride Water On Page 10 Question 1. A solution of substance X is used for white washing. (i) Name the substance X and write its formula. (ii) Write the reaction of substance X named in (i) with water. (i) We use quicklime solution for white washing. Hence, X is quicklime (CaO). (ii) Quicklime + water slaked lime + heat CaO () s + HO () l Ca(OH) ( aq) ( X) The reaction is highly exothermic and a lot of heat is produced. Question. Why is the amount of gas collected in one of the test tubes in activity 1.7 double of the amount collected in another? Name this gas. The activity 1.7 of NCERT book is electrolysis of water. During electrolysis of water, water decomposes to form hydrogen and oxygen gases. Electrolysis H O () l H ( g) + O ( g)

3 The balanced equation shows clearly that the water decomposes during electrolysis to form hydrogen and oxygen gases in the ratio : 1 by volume. Hence, the amount of hydrogen gas collected is twice or double the amount of oxygen collected. The gas is hydrogen. On Page 1 Question 1. Why does the colour of copper sulphate solution change when an iron nail is dipped in it? CuSO4 ( aq) + Fe ( s) FeSO4 ( aq) + Cu ( s) ( Blue) (Green) The above equation clearly shows that the colour of solution changes from blue to green. This is actually a displacement reaction in which a more reactive element displaces a less reactive element from its salt solution. Iron being more reactive than copper displaces copper from CuSO 4 solution (blue) to form FeSO 4 (green). Hence, change in colour is observed. Question. Give an example of double displacement reaction other than the one given in activity AgNO ( aq) + NaCl ( aq) AgCl () s + NaNO ( aq) Silver nitrate Sodium chloride Silver chloride Sodium nitrate Question. Identify the substances that are oxidised and substances that are reduced in the following reactions. (i) 4Na() s + O () s Na O() s (ii) CuO() s + H ( g) Cu () s + H O() l (i) 4 Na() s + O () g Na O() s In this reaction, Na has gained oxygen so, Na is oxidised to Na O. Here O is the oxidising agent It is itself getting reduced. Reduced 4Na () s + O() g NaO() s Oxidised Oxidised (ii) CuO ( s) + H( g) Cu ( s) + HO () l Reduced CuO is losing oxygen to form Cu ( s) so it is getting reduced and H () g is gaining this oxygen so it is getting oxidised.

4 Exercises Question 1. Which of the statements about the reactions below are incorrect? PbO () s + C () s Pb () s + CO ( g) (i) Lead is getting reduced (ii) CO is getting oxidised (iii) Carbon is getting oxidised (iv)lead oxide is getting reduced (a) (i) and (ii) (b) (i), (ii) and (iii) (c) (i) and (iii) (d) All of these (a) (i) and (ii) are incorrect. In the backward reaction, oxygen is added to lead (Pb) so Pb is getting oxidised to PbO and oxygen is removed from CO, so, it is getting reduced to C (Because addition of oxygen is oxidation and removal of oxygen is reduction.) Question. FeO + Al AlO + Fe The above reaction is an example of a (a) combination reaction (b) double displacement reaction (c) decomposition reaction (d) displacement reaction (d) In the above reaction since Al is more reactive than Fe so it displaces Fe from Fe O to form Al O. Hence, it is a displacement reaction. Question. What happens when dil. HCl is added to iron fillings? Tick the correct answer. (a) Hydrogen gas and iron chloride are produced (b) Chlorine gas and iron hydroxide are produced (c) No reaction takes place (d) Iron salt and water are produced (a) Iron being more reactive than hyrogen, displaces hydrogen from the dilute acids like dilute HCl. Thus, hydrogen gas and iron chloride are formed. Fe () s + HCl () l FeCl + H Question 4. What is a balanced chemical equation? Why should chemical equations be balanced? A chemical change is represented by a chemical equation. When the number. of atoms of different elements on reactant and product side are equal then the chemical equation is called a balanced chemical equation. It is important to balance a chemical equation because 1. to validate the law of conservation of mass which states that the mass of reactants should be equal to the mass of the products. The total mass of a system is thus conserved. This law holds true only if number. of atoms of reactants reacting together is equal to number of product atoms formed.. a balanced chemical equation tells us about the physical state of the reactants and products whether they are solid (s), liquid () l or gas () g or aqueous ( aq ).

5 . it tells us about heat changes that can take place in a chemical reaction. is the symbol of heat. Hence, it is endothermic or exothermic can be deduced from a balanced chemical equations. Question 5. Translate the following statements into chemical equation and then balance them. (a) Hydrogen gas combines with nitrogen to form ammonia. (b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide. (c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate. (d) Potassium metal reacts with water giving potassium hydroxide and hydrogen gas. (a) The symbol equation for the reaction is H + N NH The balancing of equation is done in the following steps: Step I Let us count the number of atoms of all the elements of the reactants and the products on both sides of the equation. H N 1 A simple look at the equation reveals that neither the number of H nor of N atoms are equal on both side of the equation. Step II In order to equate the number of H atoms on both sides, put coefficient before H on the reactant side and coefficient before NH on the product side. H + N NH Step III On counting, the number of N atoms on both sides of the equation, they are also the same (). This means that equation is balanced. (b) The symbol equation for the reaction is HS + O HO + SO The balancing of equation is done in the following steps: Step I Let us count the number of atoms of all the elements on both sides on the equation. H S 1 1 O

6 A simple look at the equation reveals that the number of H and S atoms are equal on both the sides. At the same time, the number of O atoms are not equal. Step II In order to equate the number of O atoms, put coefficient before O on the reactant side and coefficient before SO on the product side. HS + O HO + SO Step III O atoms are still not balanced. To achieve this, put coefficient before HO on the product side. HS + O HO + SO Step IV To balance S atoms put coefficient before HSO4 the reactant side. HS + O HO + SO Step V On inspection, the number of atoms of all the elements in both side of the equation are equal. Therefore, the equation is balanced. (c) The symbol equation for the reaction is BaCl + Al( SO4) ACl + BaSO4 The balancing of equation is done in the following steps. Step I Let us count the number of atoms of all the elements on both sides of the equation. Ba 1 1 Al 1 Cl S 1 O 1 4 A simple look at the equation reveals that only Ba atoms are equal on both the sides. The rest of the atoms are to be balanced. It is done as follows Step II In order to equate the number of Al atoms, put coefficient before AlCl on the product side. BaCl + Al ( SO ) AlCl + BaSO 4 4 Step III In order to balance Cl atoms, put coefficient before BaCl on the reactant side. BaCl + Al ( SO ) AlCl + BaSO 4 4 Step IV To balance Ba atoms, put cofficient before BaSO 4 on the product side. BaCl + Al ( SO ) AlCl + BaSO 4 4 Step V On inspection, the number S and O atoms on both sides of the equation are also found to be equal. Thus, the equation is in balanced form. (d) The symbol equation for the reaction is K + HO KOH + H The balancing of the equation is done in the following steps :

7 Step I Let us count the number of atoms of all the elements on both sides. K 1 1 H O 1 1 A simple look at the equation reveals that the number of K and O atoms on both sides of the equation are equal. At the same time, the number of H atoms are not equal. Step II To balance the number of H atoms, put coefficient before KOH on the product side and before HO on the reactant side. K + HO KOH + H Step III To balance the number of K atoms in the above equation, put coefficient before K atom on the reactant side. K + H O KOH + H Step IV On inspection, the number of atoms of all the elements are found to be equal on both sides of the equation. It is finally balanced. Question 6. Balance the following chemical equations. (a) HNO + Ca(OH) Ca(NO ) + HO (b) NaOH + HSO4 NaSO 4 + HO (c) NaCl + AgNO AgCl + NaNO (d) BaCl + H SO BaSO + HCl 4 4 (a) The symbol equation as given for the reaction is HNO + Ca(OH) Ca(NO ) + HO The balancing of the equation is done in the following steps. Step I. Let us count the number of atoms of all the elements on both the sides of the equation. H O 5 7 N 1 Ca 1 1 A simple look at the equation reveals that the number of Ca atoms are equal on both sides. Step II In order to equate the number of N atoms, put coefficient before HNO on the reactant side. HNO + Ca(OH) Ca(NO ) + HO Step III In order to equate the number of H atoms, put coefficient before HO on the product side. HNO + Ca(OH) Ca(NO ) + HO Step IV On inspection the number of O atoms on both sides of the equation is the same, i.e., 8. Therefore, the equation is balanced.

8 (b) The symbol equation as given for the reaction is NaOH+HSO4 NaSO 4 +HO Step I Let us count the number of atoms of all the elements on both sides of the equation. H O 5 5 Na 1 S 1 1 A simple look at the equation reveals that the number of O and S atoms are equal on both sides. Step II In order to equate the number of Na atoms, put coefficient before NaOH on the reactant side. NaOH+HSO4 NaSO 4 +HO Step III In order to equate the number of H atoms, put coefficient before HO on the product side. NaOH+HSO4 NaSO 4 +HO Step IV On inspection, the number of O atoms on both sides of the equation is the same i.e. 6. Therefore, the equation is balanced. (c) The symbol equation as given for the reaction is already balanced. NaCl + AgNO AgCl + NaNO (d) The symbol equation as given for the reaction is BaCl + HSO4 BaSO4 + HCl Step I Let us count the number of atoms of all the elements on both sides of the equation. Ba 1 1 H 1 O 4 4 S 1 1 Cl 1 A simple look at the equation reveals that the number of Ba, S and O atoms are equal on both the sides. Step II In order to equate the number of Cl atoms, put coefficient before HCl on the product side. BaCl` + HSO4 BaSO4 + HCl Step III On inspection the number of H atoms on both sides of the equation is the same i.e.. Therefore, the equation is balanced. Question 7. Write the balanced chemical equation for the following reactions. (a) Calcium hydroxide + carbon dioxide calcium carbonate + water (b) Zinc + silver nitrate zinc nitrate + silver

9 (c) Aluminium + copper chloride aluminium chloride + copper (d) Barium chloride + potassium sulphate barium sulphate + potassium chloride (a) Ca(OH) + CO CaCO + HO (b) Zn + AgNO Zn (NO) + Ag (c) Al + CuCl AlCl + Cu (d) BaCl + KSO4 BaSO4 + KCl Question 8. Write the balanced chemical equation for the following and identify the type of chemical reaction. (a) Potassium bromide ( aq ) + barium iodide ( aq) potassium iodide ( aq ) + barium bromide () s. (b) Zinc carbonate () s zinc oxide () s + carbon dioxide ( g) (c) Hydrogen ( g ) + chlorine ( g) hydrogen chloride ( g) (d) Magnesium () s + hydrochloric acid ( aq) magnesium chloride ( aq ) + hydrogen ( g) (a) KBr ( aq) + BaI ( aq) KI ( aq) + BaBr () s (Double displacement reaction) (Br is replaced by I and I is replaced by Br in KBr and BaI respectively.) (b) ZnCO() s ZnO() s + CO() g (Decomposition reaction (because one compound splits into two compounds) (c) H() g + Cl() g HCl( g) (Combination reaction) (because two reactants combine to form a single product) (d) Mg( s) + HCl ( aq) MgCl ( aq) + H ( g) (Displacement reaction) (because H is replaced by Mg) Question 9. What does one mean by exothermic and endothermic reactions? Give equations. On the basis of heat changes that take place during a chemical reaction we have two types of reactions exothermic and endothermic. Chemical reactions on the basis of heat changes heat is absorbed by reactants heat is released + heat heat endothermic exothermic In exothermic reactions energy of reactants is greater than energy of products. Example CH 4 + O CO + HO + Heat Methane Oxygen C + O CO + Heat

10 In endothermic reactions energy of products is greater than the energy of reactants. Example HgO( s) + Heat Hg () l + O ( g) N () g + O () g + Heat NO() g Note We write + Heat with products in exothermic reactions and +Heat with reactants in endothermic reactions. Question 10. Why is respiration considered as exothermic reaction? Explain. When we eat food we are actually consuming different nutrients like fats, proteins and carbohydrates. When the process of digestion begins in the body these complex nutrients are broken down into simpler forms like glucose and fatty acids due to the action of various enzymes. This reaction releases heat. During respiration, the air we breathe oxidises the glucose and fatty acids formed due to digestion and CO and HO are released with heat. Since, heat is released therefore respiration is exothermic in nature. C6H1O 6 + O 6CO + 6HO+Heat Glucose (from air) (Chemically Respiration) Question 11. Why are decomposition reactions called the opposite of combination reactions? Write equations for these reactions. As the name suggests when two or more substances combine together to form a new substance, it is called combination reaction. e.g., CaO + CO CaCO H + O HO In decomposition reactions just the opposite happens. One substance splits to give two or more simpler substances. Hence, it is the opposite of combination. CaCO FeSO CaO + CO 4 FeO + SO + SO Question 1. Write one equation each for decomposition reaction where energy is supplied in the form of heat, light or electricity? (i) CaCO ( s) CaO ( s)+ CO Light (ii) AgBr ( s) Ag ( s) + Br ( g) (iii) H O () l Electricity H ( g) + O ( g) Question 1. What is the difference between displacement and double displacement reactions. Write equation. I. Displacement reaction CuSO ( aq) + Fe( s) FeSO ( aq) + Cu( s) 4 4 (Blue) (Green) In this reaction since Fe is more reactive than Cu so it displaces Cu from CuSO 4 and forms FeSO 4 and sets Cu free. Hence, colour of solution changes from blue to green.

11 This is single displacement reaction. II. Double displacement reaction e.g., AgNO ( aq) + NaCl ( aq) AgCl () s + NaNO ( aq) In this reaction displacement occurs twice and mutual exchange of ions takes place between AgNO and NaCl and hence it is a double displacement reaction. AgNO + NaCl AgCl + NaNO ( aq) Question 14. In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved. The chemical equation of the displacement reaction is Cu () s + AgNO ( aq) Cu(NO ) ( aq) + Ag () s Copper Silver nitrate Copper nitrate Silver Question 15. What do you mean by precipitation reaction? Explain by giving examples. An insoluble product formed during a chemical reaction is called a precipitate. It is denoted by a downward arrow ( ). Those reactions in which reactants react to form a product which is insoluble and separates out in the form of a precipitate is called a precipitation reaction. e.g., AgNO ( aq ) + NaCl ( aq ) AgCl ( s ) + NaNO ( aq) Na SO ( aq) + BaCl ( aq) BaSO ( s) + NaCl ( aq) 4 4 Question 16. Explain the following in terms of gain or loss of oxygen with two examples each. (a) Oxidation (b) Reduction When oxygen is gained by a chemical substance, it is called oxidation. e.g., C + O CO Mg + O MgO When oxygen is lost by a chemical substance, it is called reduction. e.g., ZnO() s + C () s Zn () s + CO () g PbO + C Pb + CO Question 17. A shiny brown coloured element X on heating in air becomes black in colour. Name the element X and the black coloured compound formed. Shiny brown coloured substance is copper. It turns black on heating due to the formation of black coloured copper oxide. Cu + O CuO ( black) Copper (II) oxide

12 Hence, X = copper and compound formed is CuO (copper (II) oxide) Question 18. Why do we apply paint on iron articles? The corrosion of iron is called rusting. Rusting results in wastage of metal. Rusting occurs when (i) moisture and (ii) air are present So, to prevent the exposure of iron articles to moist air, we can apply paint so that contact is cut off and they can be prevented from rusting. In this way the surface gets protected against rusting. Question 19. Oil and fat containing food items are flushed with nitrogen. Why? Oil and fat containing food items when exposed to oxygen of air, get oxidised to form compounds that change the taste and smell of these food stuffs making them rancid. Flushing these food items with nitrogen which is an inert gas protects the food items from getting rancid. The Lays chips that you eat has nitrogen gas flushed in it to prevent rancidity. Question 0. Explain the following terms with one example of each. (a) Corrosion (b) Rancidity (a) Corrosion The process of slow degradation or eating up of metals when exposed to moist air is called corrosion. When a metal is attacked by substances around it such as moisture and acids etc., it is said to corrode and the process is called corrosion. eg.., the black coating on silver and the green coating on copper are other examples of corrosion. Silver develops black coat due to hydrogen sulphide of air. Ag + HS AgS + H (b) Rancidity When fats and oils are oxidised, they become rancid and their smell and taste get changed. Usually substances which prevent oxidation (antioxidants) are added to foods containing fats and oils. Keeping food in air tight containers helps to slow down oxidation. This is the reason why chips which we eat are usually flushed with nitrogen gas to prevent rancidity.

Chapter 1 Chemical Reactions & Equations

CBSE Class 10th NCERT Science Chapter 1 Chemical Reactions & Equations Intext Questions On Page 6 Question 1: Why should a magnesium ribbon be cleaned before burning in air? Magnesium is an extremely reactive