In-text Questions (Page 10)<\/span><\/p>\nQuestion 1.
\nA solution of a substance \u2018X is used for white-washing:
\n(i) Name the substance ‘X’ and write its formula.
\n(ii) Write the reaction of the substance ‘X’ named in (i) above with water.
\nAnswer:
\n(i) The substance ‘X’ is quick lime or calcium oxide.
\n<\/p>\n
Question 2.
\nWhy is the amount of gas collected in one of the test tubes in activity 1.7 double of the amount collected in the other ? Name this gas.
\nAnswer:
\nWhen electricity is passed through the water in presence of an acid., water is electroysed and produces hydrogen and oxygen gas.
\n\\(2 \\mathrm{H}_{2} \\mathrm{O}(l) \\stackrel{\\text { Electrolysis }}{\\longrightarrow} 2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g)\\)
\nIt is clear from the above equation that 2 moles .water undergoes electrolysis and produce 2 moles of hydrogen and one mole of oxygen in the given set of conditions of temperature and pressure because the volume of a gas is directly proportional to its no. of moles so the amount of one gas is doubled than the other.
\nThe double amount gas is hydrogen.<\/p>\n
<\/p>\n
In-text Questions (Page 13)<\/span><\/p>\nQuestion 1.
\nWhy does the colour of copper sulphate solution change when an iron nail is dipped in it?
\nAnswer:
\nIron is more reactive that copper so it displaces copper metal from copper sulphate solution and produces sulphate of iron.
\n<\/p>\n
Question 2.
\nGive an example of a double displacement reaction other than the one given in Activity 1.10.
\nAnswer:
\n<\/p>\n
Question 3.
\nIdentify the substances that are oxidised and the substances that are reduced in the
\nfollowing reactions.
\n(i) Na(s) + O2<\/sub>(g) \u2192 2Na2<\/sub>O(s)
\n(ii) CuO(s) + H2<\/sub> (g) \u2192 CU(s) + H2<\/sub>O(l)
\nAnswer:
\n(i) 2 Na (s) + O2<\/sub> (g) \u2192 2 NO2<\/sub>O (s)
\nNa (s) \u2192 Oxidised
\nO2<\/sub>(g) \u2192 Reduced
\n(ii) CuO (s) + H2<\/sub>(g) \u2192 Cu (S) + H2<\/sub>O (l)
\nCuO (s) \u2192 Reduced
\nH2 <\/sub>(g) \u2192 Oxidised<\/p>\nClass 10 Science Chapter 1 Chemical Reactions and Equations Textbook Questions and Answers<\/h3>\n
Page No. 14<\/span><\/p>\nQuestion 1.
\nWhich of the statements about the reaction below are incorrect?
\n2PbO(s) + C(s) \u2192 2Pb(s) + CO2<\/sub>(g)
\n(a) Lead is getting reduced.
\n(b) Carbon dioxide is getting oxidised.
\n(c) Carbon is getting oxidised.
\n(d) Lead oxide is getting reduced.
\n(i) (a) and (b)
\n(ii) (a) and (c)
\n(iii) (d), (b) and (c)
\n(iv) all
\nAnswer:
\n(i) a and b<\/p>\nQuestion 2.
\nFe2<\/sub>O3<\/sub> + 2Al \u2192 Al2<\/sub>O3<\/sub> + 2Fe
\nThe above reaction is an example of a
\n(a) combination reaction.
\n(b) double displacement reaction.
\n(c) decomposition reaction.
\n(d) displacement reaction.
\nAnswer:
\n(d) displacement reaction.<\/p>\nQuestion 3.
\nWhat happens when dilute hydrochloric add is added to iron fillings? Tick the correct answer.
\n(a) Hydrogen gas and iron chloride are produced.
\n(b) Chlorine gas and iron hydroxide are produced.
\n(c) No reaction takes place.
\n(d) Iron salt and water are produced.
\nAnswer:
\n(a) Hydrogen gas and iron chloride are produced.<\/p>\n
<\/p>\n
Question 4.
\nWhat is a balanced chemical equation? Why should chemical equations be balanced?
\nAnswer:
\nA chemical equation is said to be balanced when it has the same no. of atoms of different elements in both the sides, i. e., reactant and products sides.
\ne.g. H2<\/sub> + O2<\/sub> \u2192 H2<\/sub>O
\nThe above equation is not a balanced equation because it does not have the equal no. of atoms of oxygen in both sides.
\nBut 2H2<\/sub> + O2<\/sub> \u2192 2 H2<\/sub>O is a balanced equation because it does not have the equal no. atoms of different elements in both the sides.<\/p>\nA chemical equation should be balanced because during a chemical change the no. of atoms of each element remain same. In other words the law of concervation of mass i.ethe total mass of all the products of a chemical reaction has to equal to the total mass of all reactants. And it is only possible when a chemical equation is balanced.<\/p>\n
Question 5.
\nTranslate the following statements into chemical equations and then balance them.
\n(a) Hydrogen gas combines with nitrogen to form ammonia.
\n(b) Hydrogen sulphide gas bums in air to give water and sulpur dioxide.
\n(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
\n(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
\nAnswer:
\n(a) H2<\/sub> (g) + N2<\/sub>(g) \u2192 NH3<\/sub> (g)
\nStep I. Look more closely the number of atoms of different elements present in the unbalanced chemical equation.<\/p>\n\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n |
\nH<\/td>\n | 2<\/td>\n | 3<\/td>\n<\/tr>\n |
\nN<\/td>\n | 2<\/td>\n | 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Step II. It is often convenient to start with the compound that contains maximum number of atoms, whether a reactant or a product. So we select NH3 and the element hydrogen in it. There are three hydrogen atoms on the right and only two hydrogen atoms on the left.<\/p>\n \n\n\nAtoms of Hydrogen<\/td>\n | In reactants<\/td>\n | In Products<\/td>\n<\/tr>\n | \nInitial<\/td>\n | 2 (in H2<\/sub>)<\/td>\n3 (in NH3<\/sub>)<\/td>\n<\/tr>\n\nTo balance<\/td>\n | 2 \u00d7 3<\/td>\n | 3 \u00d7 2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Now the partly balanced equation becomes as follows : \n3H2<\/sub> + N2<\/sub> \u2192 2NH3<\/sub><\/p>\nStep III. But nitrogen in the above equation is automatically balanced. So the balanced equation will be \n3 H2<\/sub> + N2<\/sub> \u2192 NH3<\/sub> \nThe above equation balanced because it contains the equal no. of atoms of hydrogen and nitrogen on both sides. \n(b) H2<\/sub>S + O2<\/sub> \u2192 H2<\/sub>O + H2<\/sub>O \nThe above equation can be balanced by algebraic sum method.<\/p>\nLet ‘a’, ‘b’, \u2018c’ and ‘d’ are the no. of molecules of ‘H2<\/sub>S’, ‘O2<\/sub>‘, ‘H2<\/sub>O’ and ‘SO2<\/sub>‘ respectively in the balanced chemical equation. So the above equation can be written as \na H2<\/sub>S + b O2<\/sub> \u2192 c H2<\/sub>O + d SO2<\/sub> \nNow in a balanced chemical equation. \nL.H.S. R.H.S. \nHydrogen 2 a= 2c …(i) \nSulphur a = d …(ii) \nOxygen 2b = c + 2d ….(iii) \nLet a = 1 …..(iv)<\/p>\nSo from equations (i) and (iv) \n2 \u00d7 1 = 2 c \n\u21d2 c = 1 \nFrom equation (iii) \n2b = c + 2d \n2b = 1 + 2 \u00d7 1 \n\u21d2 2b = 3 \n\u21d2 b = \\(\\frac {3}{2}\\) \nNow by putting the value of a, b, c and d in the given equation, we get \n1 \u00d7 H2<\/sub>S + \\(\\frac {3}{2}\\)O2<\/sub> \u2192 1 \u00d7 H2<\/sub>O + 1 \u00d7 SO2<\/sub> \nOR 2 H2<\/sub>S + 3O2<\/sub> \u2192 2 H2<\/sub>O + 2SO2<\/sub><\/p>\n(c) \n<\/p>\n Step I:<\/p>\n \n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n | \nBa<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n | \nCl<\/td>\n | 2<\/td>\n | 3<\/td>\n<\/tr>\n | \nAl<\/td>\n | 2<\/td>\n | 1<\/td>\n<\/tr>\n | \nS<\/td>\n | 3<\/td>\n | 1<\/td>\n<\/tr>\n | \nO<\/td>\n | 12<\/td>\n | 4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Step II: Now start from Al2<\/sub> (SO4<\/sub>)3<\/sub> and balance oxygen first. The simple ratio between the atoms of oxygen in reactants to the product will be 12 : 4 or 3 : 1. So multiplying Al2<\/sub> (SO4<\/sub>)3<\/sub> by ‘1’ ‘BaCl2<\/sub>‘ by ‘3’, we get \nBaCl2 + Al2 (SO4)3 \u2192 AlCl3 + 3BaSO4<\/p>\nStep III: Now balance ‘Al’ be multiplying ‘2’ in the product side (i.e. in AlCl3<\/sub>) \nBaCl2<\/sub> + Al2<\/sub> (SO4<\/sub>)3<\/sub> \u2192 2 AlCl3<\/sub> + 3 BaSO4<\/sub><\/p>\nStep IV : Balance barium by multiplying ‘3’ in the reactant to BaCl2<\/sub>. \n3 BaCl2<\/sub> + Al2<\/sub> (SO4<\/sub>)3<\/sub> \u2192 2 AlCl3<\/sub> + 3 BaSO4<\/sub> \n‘Cl’ is automatically balanced in the above equation.<\/p>\n(d) K + H2<\/sub>O \u2192 KOH + H2 \n<\/sub>Step I.<\/p>\n\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n | \nK<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n | \nH<\/td>\n | 2<\/td>\n | 3<\/td>\n<\/tr>\n | \nO<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Step II: Now select ‘KOH’ in the reactant side because it contains the maximum no. of elements. \nK + 2H2<\/sub>O \u2192 2 KOH + H2<\/sub><\/p>\nStep III: Balance potassium metal in the above partially balanced equation by multiplying ‘2’ in the potassium i.e. reactant side. \n2K + 2H2<\/sub>O \u2192 2KOH + H2<\/sub><\/p>\n<\/p>\n Question 6. \nBalance the following chemical equations. \n(a) HNO3<\/sub> + Ca(OH)2<\/sub> \u2192 Ca(NO3<\/sub>)2<\/sub> + H2<\/sub>O \n(b) NaOH + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + H2<\/sub>O \n(c) NaCl + AgNO3<\/sub> \u2192 AgCl + NaNO3<\/sub> \n(d) BaCl2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub> + HCl \nAnswer: \n(a) \nStep I.<\/p>\n\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n | \nH<\/td>\n | 3<\/td>\n | 2<\/td>\n<\/tr>\n | \nN<\/td>\n | 1<\/td>\n | 2<\/td>\n<\/tr>\n | \nO<\/td>\n | 5<\/td>\n | 7<\/td>\n<\/tr>\n | \nCa<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Step II. Select Ca (NO3<\/sub>)2<\/sub> because it contains the maximum no. of elements. To balance oxygen multiply Ca (OH)2 <\/sub>by two, we get \nHNO3<\/sub>+ 2Ca (OH)2<\/sub> \u2192 Ca(NO3<\/sub>)2<\/sub> + H2<\/sub>O<\/p>\nStep III. Now balance calcium atoms by multiplying two in reactants sides. We get \nHNO3<\/sub> + 2 Ca (OH)2<\/sub> \u2192 Ca (NO3<\/sub>)2<\/sub> + H2<\/sub>O<\/p>\nStep IV : Balance nitrogen in the above partially balanced equation, multiply HNO3<\/sub> by ‘4’. We get \n4 HNO3<\/sub> + 2 Ca (OH)2<\/sub> \u2192 2 Ca (NO3<\/sub>)2<\/sub> + H2<\/sub>O<\/p>\nStep V : Balance hydrogen by multiplying ‘4’ in the product side. We get, \nHNO3<\/sub> + 2 Ca (OH)2<\/sub> \u2192 2 Ca (NO3<\/sub>)2<\/sub> + 4H2<\/sub>O \nThis is a balanced chemical equation.<\/p>\n(b) NaOH + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + H2<\/sub>O \nStep I.<\/p>\n\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n | \nNa<\/td>\n | 1<\/td>\n | 2<\/td>\n<\/tr>\n | \nO<\/td>\n | 5<\/td>\n | 5<\/td>\n<\/tr>\n | \nH<\/td>\n | 3<\/td>\n | 2<\/td>\n<\/tr>\n | \nS<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Step II: Balance sodium in the above equation by multiplying ‘2’ in NaOH. \n2 NaOH + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + H2<\/sub>O<\/p>\nStep III: Balance hydrogen by multiplying ‘2’ in H2<\/sub>O. \n2 NaOH + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + 2 H2<\/sub>O \nThis is a balanced equation.<\/p>\n(c) NaCl + AgNO3<\/sub> \u2192 AgCl + NaNO3<\/sub> \nStep I.<\/p>\n\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n | \nNa<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n | \nCl<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n | \nAg<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n | \nN<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n | \nO<\/td>\n | 3<\/td>\n | 3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n The above equation contains equal no. of atoms of different elements in reactants and products sides.<\/p>\n (d) BaCl2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub> + HCl \nStep I.<\/p>\n\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n | \nBa<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n | \nCl<\/td>\n | 2<\/td>\n | 1<\/td>\n<\/tr>\n | \nH<\/td>\n | 2<\/td>\n | 1<\/td>\n<\/tr>\n | \nS<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n | \nO<\/td>\n | 4<\/td>\n | 4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Step II : Now balance chlorine atoms in the above equation by multiplying ‘2’ in the product side to HCl \nBaCl2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub> + 2HCl \nThis is a balanced chemical equation.<\/p>\n<\/p>\n Question 7. \nWrite the balanced chemical equations for the following reactions. \n(a) Calcium hydroxide + Carbon dioxide \u2192 Calcium carbonate + Water \n(b) Zinc + Silver nitrate \u2192 Zinc nitrate + Silver \n(c) Aluminium + Copper chloride \u2192 Aluminium chloride + Copper \n(d) Barium chloride + Potassium sulphate \u2192 Barium sulphate + Potassium chloride. \nAnswer: \n(a) Ca (OH2<\/sub>) + CO2<\/sub> \u2192 CaCO3<\/sub> + H2<\/sub>O \nStep I.<\/p>\n\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n | \nCa<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n | \nO<\/td>\n | 4<\/td>\n | 4<\/td>\n<\/tr>\n | \nH<\/td>\n | 2<\/td>\n | 2<\/td>\n<\/tr>\n | \nC<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n The above equation is balanced.<\/p>\n (b) Zn + AgNO3<\/sub> \u2192 Zn (NO3<\/sub>)2<\/sub> + Ag \nStep I.<\/p>\n\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n | \nZn<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n | \nAg<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n | \nN<\/td>\n | 1<\/td>\n | 2<\/td>\n<\/tr>\n | \nO<\/td>\n | 3<\/td>\n | 6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Step II: Balance oxygen by multiplying ‘2’ in AgNO3<\/sub>, we get \nZn + 2 AgNO3<\/sub> \u2192 Zn (NO3<\/sub>)2<\/sub> + Ag<\/p>\n | | | | | | | | | | | |