Equations and Stoichiometry
Equations and Stoichiometry Chemistry 100
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This 4 page Bundle was uploaded by firstname.lastname@example.org Notetaker on Monday October 26, 2015. The Bundle belongs to Chemistry 100 at Indiana University of Pennsylvania taught by Dr. Ronald See in Fall 2015. Since its upload, it has received 45 views. For similar materials see Preparatory Chemistry in Chemistry at Indiana University of Pennsylvania.
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Date Created: 10/26/15
October 1October 13 Equations and Stoichiometry 1 Balancing Chemical Equations a Shorthand to describe a chemical reaction b Reactant compounds on left of the arrow c Products compounds on the right of the arrow i Formed by the reaction 2 States of Matter a Four general states i Gas g ii Liquid I iii Solid 5 iv Aqueous aq 1 Compound is dissolved in water 3 Examples of Chemical Equations a H2g02gH20 I i Hydrogen gas and oxygen gas combine to make liquid water b Add coef cients to equations I H202H20 i 2 is the coef cient ii No coef cient1 4 Balancing Equations a Need equal number of each element on each side of the H i Balance each element one by one b Steps to Balance i 1 Start with the element that appears only once on each side and has a subscript change ii 2 Use least common multiple LCM to assign initial coef cients iii 3 Repeat Steps 1 and 2 for the next elementsif coef cient from 1St element will not work double them 1 2142 iv 4 Leave elements that are given as single atoms metals for last 1 Easiest to do v 5 When balanced check coef cients to see if all are divisible by the same number 1 Get coef cients to lowest term 5 Examples of Balancing Equations a Al302 2Al203 i 6LCM b P4Cl24 PCl3gt P46Clj 4PC3 c P406H20H3PO3 i P4066H20j4H3PO3 i Balance polyatomic ions as 1 single element 6 Chemical Equations from Description of Reaction a Elements form the way elements are found in nature i Nonmetals Group 18 always single atoms ii C and Si are also single atoms i 02 Oxygen gas c Other nonmetals have various element forms d Metals are all single atoms i Fes Ks 7 Procedure a Steps i 1 Write formula from names to give unbalanced reaction ii 2 Balance reaction include coefficient b Examples i Aqueous solutions of calcium chloride and potassium carbonate react to form solid calcium carbonate and aqueous potassium chloride 1 Calcium chloride CaC2 aq 2 Potassium carbonate K2CO3 3 Calcium carbonate Ca2C02393leaCO3s 4 Potassium chloride KCl39KC aq 5 CaCl2K2CO3jCaCO32KC ii Sulfur trioxide gas reacts with chlorine gas to form SOzC2 oxygen gas 1 Sulfur trioxideSO3 2 Oxygen gasD 02 3 SO3 gCl2 gIISOZC2g 02 9 4 25032C2D2502C202 8 Simple Stoichiometry Calculations a Steps for nding amount in a chemical reaction i 1 Write balanced equation ii 2 Convert given compound from grams to moles divide by Molar Mass iii Convert moles of given compound in moles of answer compound using stoic ratio 1 iv Convert moles of answer compound to grams 1 Multiple by Molar Mass of answer compound b Can be done with any 2 compound in balanced equation i Reactant and product ii 2 reactants iii 2 products c Examples i If 239g N204 is consumed how much water can be formed 1 H2044H2D4H20N2 2 Molar Mass of N204239g 1 mol920g00260mo N204 3 Ratio 4 mol H20 1 mol N204 x mol H2000260 mol N204 a 4 mol H20 00260mol N204 1 mol N204X mol H20 0104 mol H20 4 Molar Mass H20 180 gmol 5 0104 mol H20 180 g1 mo187 g H20 9 Limiting Reagent a Simple stoic only one mass is given other compounds are assumed to be in abundant supply in excess i Limitind readent runs out rst smaller moles c Steps i 1 Carry out 2 393rCI steps of stoic on both reactants ii 2 Compare moles of answer compound reactant smaller number of moles is the limiting reagent iii 3 Final step with only the moles from the limiting reagent d Example i How much KCO3 can be formed from 550g KCI and 275g 02 1 Balanced equation 2 KCI 302D 2KCO3 2 Molar Mass of KCI 746 gmol a 550g 1 mol746g 00737 mol KCI b 2 mol KCIO32 mol KCI x mol KCIO3 00737 mol KCI c 11 ratio d X 00737 mol KCO3 3 Molar Mass of 02 320 gmol a 275 g 1 mol320g 00859 mol 02 b 2 mol KCIO3 3 mol 02x mol KCIO3 00859 mol 02 c X 00573 mol KCO3 4 Molar Mass of KCIO3 1226 gmol a 00573 mol 1226g 1 mol 7 01 g KCO3 10 Percent Yields a Theoretical vield results of stoic i How much you could possibly make ii To actually reach TY is almost impossible c Example i 2 CH4202 CH3COOH 2 H20 1 yield 715 2 How much CH3COOH will be formed from 100 g of CH4 ii Molar Mass of CH4 16 gmol 1 100 g CH4 1 mol16 g 0625 mol CH4 2 1 mol CH3COOH2 mol CH4xO625 mol CH4 a X0313 mol CH3COOH 3 0313mol 600g1 mo 188 g Theoretical Yield iii 715 actual100188g 1 Actual 715188100 iv Actual yied 134 g
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