STOICHIOMETRY TUTORIAL 1 1 (1--3) General Approach For Problem Solving: 1. Clearly identify the Goal or Goals and the UNITS involved. (What are you trying to do?). Determine what is given and the UNITS. 3. Use conversion factors (which are really ratios) and their UNITS to CONVERT what is given into what is desired. Sample problem for general problem solving. Sam has entered into a 10 mile marathon. Use ALL of the following conversions (ratios) to determine how many inches there are in the race. 580 ft = 1 mile; 1 inches = 1 ft 1. What is the goal and what units are needed?. What is given and its units? Goal = inches 10 miles 3. Convert using factors (ratios). Units match 10 miles = 633600 inches Given Convert Goal 3 Menu 3
Sample problem # on problem solving. A car is traveling at a speed of 45 miles per hr (45 miles/hr). Determine its speed in kilometers per second using the following conversion factors (ratios). 1 mile = 580 ft; 1 ft = 1 in; 1 inch =.54 cm; km = 1 x 10 3 ; cm = 1 x 10 - m; 1 hr =60 min; 1 min = 60 s Given Goal = 0.00 km s c cancels c m remains This is the same as putting k over k Units Match! 4 4 Converting grams to moles. Determine how many moles there are in 5.17 grams of Fe(C 5 H 5 ). Given units match Goal 5.17 g Fe(C 5 H 5 ) = 0.078 moles Fe(C 5 H 5 ) Use the molar mass to convert grams to moles. Fe(C 5 H 5 ) x 5 x 1.001 = 10.01 x 5 x 1.011 = 10.11 1 x 55.85 = 55.85 5 5 Stoichiometry (more working with ratios) Ratios are found within a chemical equation. HCl + 1Ba(OH) H O + 1 BaCl coefficients give MOLAR RATIOS moles of HCl react with 1 mole of Ba(OH) to form moles of H O and 1 mole of BaCl 6 6
Mole Mole Conversions When N O 5 is heated, it decomposes: a. How many moles of NO can be produced from 4.3 moles of N O 5? 4.3 mol? mol Units match 4.3 mol N O 5 = 8.6 moles NO b. How many moles of O can be produced from 4.3 moles of N O 5? 4.3 mol? mol 4.3 mol N O 5 =. mole O 7 7 gram mole and gram gram conversions When N O 5 is heated, it decomposes: a. How many moles of N O 5 were used if 10g of NO were produced?? moles 10g Units match 10 g NO =.8 moles N O 5 b. How many grams of N O 5 are needed to produce 75.0 grams of O? 75.0 g? grams 75.0 g O = 506 grams N O 5 8 8 Gram to Gram Conversions Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + 6 HCl(aq) AlCl 3 (aq) + 3H (g) First write a balanced equation. 9 9
Gram to Gram Conversions Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + 6 HCl(aq) AlCl 3 (aq) + 3H (g) 3.45 g? grams Now let s get organized. Write the information below the substances. 10 10 gram to gram conversions Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? 3.45 g Al Al(s) + 6 HCl(aq) AlCl 3 (aq) + 3H (g) 3.45 g? grams Units match = 17.0 g AlCl 3 We Now Let s must use work the always the molar problem. convert mass ratio. to convert moles. to grams. 11 11 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide,, is used in rebreathing gas masks to generate oxygen. a. How many moles of O can be produced from 0.15 mol and 0.10 mol H O? First copy down Now the place the BALANCED numerical the equation! information below the compounds. 1 1
Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide,, is used in rebreathing gas masks to generate oxygen. a. How many moles of O can be produced from 0.15 mol and 0.10 mol H O? 0.15 mol 0.10 mol? moles Two starting amounts? Where do we start? Hide one 13 13 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide,, is used in rebreathing gas masks to generate oxygen. a. How many moles of O can be produced from 0.15 mol and 0.10 mol H O? 0.15 mol 0.10 mol? moles Hide 0.15 mol KO = 0.115 mol O 14 14 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide,, is used in rebreathing gas masks to generate oxygen. a. How many moles of O can be produced from 0.15 mol and 0.10 mol H O? 0.15 Hide mol 0.10 mol? moles 0.15 mol KO = 0.115 mol O H O 0.10 mol H O = 0.150 mol O 15 15
Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide,, is used in rebreathing gas masks to generate oxygen. a. How many moles of O can be produced from 0.15 mol and 0.10 mol H O? Determine the limiting reactant. 0.15 mol 0.10 mol? moles 0.15 mol KO = 0.115 mol O 0.10 mol H O H O H O = excess (XS) reactant! It was limited by the amount of. = 0.150 mol O What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant? 16 16 Theoretical yield vs. Actual yield Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 1.3 grams of product were recovered. Determine the % yield. Theoretical yield = 19.5 g based on limiting reactant Actual yield = 1.3 g experimentally recovered 17 17 Limiting/Excess Reactant Problem with % Yield If a reaction vessel contains 10.0 g of and 47.0 g of H O, how many grams of O can be produced? 10.0 g 47.0 g? g Hide one 10.0 g = 40.51 g O 18 18
Limiting/Excess Reactant Problem with % Yield If a reaction vessel contains 10.0 g of and 47.0 g of H O, how many grams of O can be produced? 10.0 g 47.0 g? g Hide 10.0 g = 40.51 g O H O 47.0 g H O = 15.3 g O Question if only 35. g of O were recovered, what was the percent yield? 19 19 If a reaction vessel contains 10.0 g of and 47.0 g of H O, how many grams of O can be produced? 10.0 g 47.0 g? g 10.0 g = 40.51 g O H O 47.0 g H O Determine how many grams of Water were left over. = 15.3 g O The Difference between the above amounts is directly RELATED to the XS H O. 15.3-40.51 = 84.79 g of O that could have been formed from the XS water. 84.79 g O = 31.83 g XS H O 0 0