Solve this:

Solve this: Here. methane and dioxygen are called reactants and carbon dioxide and water are called prulucts. Note that all the reactants and the products are gases in the above reaction and this has been indicated by letter (g) in the brackets next to its fortnula. Similarly. in the case Of solids and liquids. (s) and (l) are written respec tively. "Itie coemcients 2 for Oa and Hao are called stoichiometric coefficients. Similarly the coefficient for CH and CO is one in each case. They represent the number Of molecules (and moles as well) taking part in the reaction or formed in the reaction. Thus. according to the above chemical reaction. One mok of CH (g) reacts With two moles Of 02(g) to give one mole Of CCMg) and two moles Of H 20(g) One molecule Of CH (g) reacts with 2 molecules Of 02(g) to give one molecule Of C02(g) and 2 molecules Of H20(g) 22.7 L reacts with 45.4 L of 02 (g) to give 22.7 L ofC02 (g) and 45.4 16 gofCH4 (g) reacts with 2x32 gof02 (g) to give 44 g Of C02 (g) and 2xIS g Of H20 From these relationships. the given data can be interconverted as follows : mass moles* nu of molecules = Density Volume Problem 1.3 Calculate the amount Of water (g) produced by the combustion Of 16 g Of methane. Solution The balanced equation for combustion Of methane is : +202 (g) +2H20 (g) (i) 16 g Of CH. con-esponds to one mole. (ii) From the above equation, I mol Of CH. (k) gives 2 mol of H20 (g). 2 mol ofwater 2 x (2+16) 18=36g 18g 1 mol H20- 18 g H20 Imol H20 1 18g H20 Hence 2 mol H20 x Imol H20 -2 x 18gHzo-36gHß) Problem 1.4 HOW many moles Of methane are to produce 22 g C02 (g) after combustion? Soluti on According to the chemical equation, (g) +202 C02 (g) +21120 (g) u44gc02 (g) is obtained from 16 g CH. I I mol C02(g) iS obtained from I mol Of CH. (gil mole of COz (g) 1 mol C02 (g) 22 g 44 0.5 Hence. O. 5 mol C02 @would be obtained from 0.5 mol CH. (g) or 0.5 mol Of CH. (g) would be required to produce 22 g I. I O. I Limiting Reagent Many a time. the reactions are carried out when the reactants are not present in the amounts as required by a balanced chemical reaction. In such situations. one reactant is in excess over the Other. •me reactant Which is present in the lesser amount gets consumed after sometime and after that no further reaction takes place whatever be the amount Of the other reactant present. Hence. the reactant Which gets consumed. limits the amount Of product formed and is. therefore. called the limiting reagent. In performing stoichiometric calculations. this aspect is also to be kept in mind.