hi guys welcome to igcs study bu where you can revise chemistry topics from the Cambridge igcc syllabus if you are enjoying our video so far please don't forget to hit the like button and subscribe to our channel in this video you are going to learn part two of topic three stochiometry word equations are like written descriptions of chemical reactions you describe what's happening using words for example sodium hydroxide and hydrochloric acid react to form sodium chloride and water sodium hydroxide and hydrochloric acid are the reactants in this reaction reactants are the substances you start with in a chemical reaction the products or the substances that result from the reaction in this example are sodium chloride and water state symbols tell us the physical state of the substances involved s means solid l means a liquid g means gas and AQ means Aquas it indicates that a substance is dissolved in water so for our previous example these will be the state symbols in symbol equations instead of words we use chemical symbols or formulas like NaOH for sodium hydroxide and H2O for water so the equation for the reaction above becomes NaOH plus HC to give n+ H2O let's look at some guidelines for naming compounds if the compound has a metal and a non-metal the metal comes first the non-metal ends with ID D for example pottassium bromide if the compound consists of two non-metals and one of them is hydrogen then hydrogen comes first for example hydrogen bromide if the compound consists of two non-metals the lower group number element comes first and if it's a compound with a metal and a common ion group the metal comes first these are the molecules that commonly occur with two atoms and we show them using the format for example H2 moving on to ionic equations some reactions involve ions or charged atoms we use special equations to show these let's start with a balanced equation taking our same example let's split up the aquous compounds into ions because in water ionic compounds break apart into their individual ions separating them into the ions that originally composed them so that gives na+ plus o minus plus H+ plus c CL minus to give na+ plus CL minus and H2O cancel out the ions that are the same on either side then we are left with this ionic equation h++ o minus to give H2O deducing symbol equations this means figuring out the right symbol equation for a reaction when you are given information about it so example one calcium reacts with chlorine gas to form calcium chloride this is written as ca+ cl2 the two in cl2 means there are two chlorine atoms in the diatomic molecule calcium is a Group Two element with a valency of Plus + 2 and chlorine has a valency of minus1 the valencies are crisscrossed to derive the formula of ca2 example two Write the balanced chemical equation for the reaction between sulfuric acid and pottassium hydroxide sulfuric acid dissociates into H+ and S so 4 2 minus ions and potassium hydroxide dissociates into k+ and O minus ions the H+ ions from sulfuric acid react with the oh minus ions from potassium hydroxide to form water the remaining ions k+ and S so42 minus combined to form pottassium sulfate or K2 so4 now this equation is not balanced balancing a chemical equation means making sure there are the same numbers of each type of atom on both sides you use coefficients or numbers in front of chemicals to achieve this let's compare the number of atoms on the left hand side to the right hand side of the equation so we have three hes on the left side and on the right hand side only two h's so let's just try putting a two in front of the k for now balancing equations is all about trial and error let's forget about the H for a while and check if the S4s are balanced on each side and it looks like it is if you come across common ion groups like s so42 minus that doesn't change from one side to the other just count it as one thing not all the separate parts now K pottassium we have two k on the left hand side as well as two on the right hand side other than the O's in the sulfates there are two o's on the left hand side because we added a two in front of the KO o h but only one o on the right hand side so let's try adding a two in front of the H2O and see if that helps to balance this so now the O's are balanced there are four hes on the right hand side because 2 * 2 is 4 the left hand side also has 4 hes looks like our equation is now balanced you must always double check after using this trial and error method now you might be getting the hang of it in this example you have been asked for the balanced chemical equation for the oxidation of ion when it reacts with oxygen to form ion 3 oxide ion can exhibit different oxidation States in this case it's mentioned three in the ion 3 oxide so we know it forms a fe3+ ion so let's see how the formula is derived crisscrossing the valencies gives you Fe2O3 now let's balance this equation we have one Ion on the left and two on the right to balance let's put a two in front of Fe on the left now let's check oxygen we have two oxygen at atoms on the left but we have three on the right let's put a three in front of the O2 on the left now there are six O's on the left side and on the right side if we put a two in front of the Fe2O3 then we will get six oxygen atoms on the right side as well let's double check we have two ion atoms on the left side and four Ion at atoms on the right side so let's change the number in front of the ion to four now the equation is balanced in this example calcium oxide reacts with nitric acid to produce calcium nitrate and water notice how the NO3 is enclosed in Brackets showing that the two belongs to the whole nitrate ion to balance this equation we simply add a two in front of the hno3 on the left side in the next equation there are three products that concludes part two of topic three stochiometry are you enjoying our videos are they helping you here's a way you can show your appreciation and 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