Stoichiometry Part II

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Stoichiometry Part II Name Section 1. Given an equation C H 6 (g) + 7O (g) 4CO (g)+ 6H O(g) How many mol of CO will be formed by the complete combustion of 6.6 mol C H 6? The question can be interpreted into the ICE Table as we are looking for the amount of CO given 6.6 mol C H 6 and an excess of O. The amounts of CO (g) and H O(g) are assumed to be 0 initially. C H 6 (g) + 7O (g) 4CO (g) + 6H O(g) Initial 6.6 mol excess 0 0 The question states that the 6.6 mol C H 6 are completely combusted then we know that 6.6 moles C H 6 will completely react. This places 6.6 mol in the row for C H 6. Once we have an entry in the row we can use the stoichiometric coefficients to determine the other entries. C H 6 (g) + 7O (g) 4CO (g) + 6H O(g) Initial 6.6 mol excess 0 0-6.6 mol Calculate the moles of CO formed 6.6 mol C H 6 4 mol CO mol C H = 13. mol CO 6 C H 6 (g) + 7O (g) 4CO (g) + 6H O(g) Initial 6.6 mol excess 0 0-6.6 mol 13. mol 13. mol Since there were no mol of CO initially,then the amount of CO is 0 + 13. mol = 13. mol We can also determine the remaining entries in the row as 6.6 mol C H 6 6 mol H O mol C H = 19.8 mol H O 6 6.6 mol C H 6 7 mol O mol C H = 3.1 mol O 6

C H 6 (g) + 7O (g) 4CO (g) + 6H O(g) Initial 6.6 mol excess 0 0-6.6 mol 3.1 mol 13. mol 19.8 mol 0 mol excess 13. mol 19.8 mol How many moles of C H 6, assuming excess oxygen, are required to form 3.7 mol H O? The question can be interpreted into the ICE Table as we are looking for the Initial amount of C H 6 given 3.7 mol H O are formed ( amount) and an excess of O. The amounts of CO (g) and H O(g) are assumed to be 0 initially. C H 6 (g) + 7O (g) 4CO (g) + 6H O(g) Initial excess 0 0 3.7 mol Since there was no H O present initially (and no CO ) we can determine that 3.7 mol H O should be in the row. C H 6 (g) + 7O (g) 4CO (g) + 6H O(g) Initial excess 0 0 +3.7 mol 3.7 mol Once we have an entry in the row we can use the stoichiometric coefficients to determine the other entries. 3.7 mol H O mol C H 6 6 mol H O = 1. mol C H 6 C H 6 (g) + 7O (g) 4CO (g) + 6H O(g) Initial excess 0 0-1. mol +3.7 mol 3.7 mol So there must be 1. mol C H 6 present initially for that much to react. Again we can use the stoichiometric coefficients to determine the other entries in the row and the row. 3.7 mol H O 4 mol CO 6 mol H O =.47 mol C H 6

3.7 mol H O 7 mol O 6 mol H O = 4.3 mol C H 6 C H 6 (g) + 7O (g) 4CO (g) + 6H O(g) Initial 1. mol excess 0 0-1. mol -4.3 mol +.47 mol +3.7 mol 0 mol excess.47 mol 3.7 mol. Determine the amount of iodine produced when 145 g of KI react with excess copper (II) chloride. CuCl (s) + 4KI(s) CuI(s) + 4KCl(s) + I (s) The question can be interpreted into the ICE Table as we are looking for the amount of I given 145 g KI and an excess of CuCl. The amounts of CuI, KCl and I (g) are assumed to be 0 initially. The ICE Table entries must be in moles so we must first convert grams of KI to moles of KI using the molar mass of KI. 145 g KI 1 mol KI 166 g = 0.873 mol KI reacting CuCl (s) + 4KI(s) CuI(s) + 4KCl(s) + I (s) Initial excess 0.873 mol 0 0 0-0.873 mol So we know that 0.873 mol of KI are reacting we can use the stoichiometric coefficients to calculate how much I forms 0.873 mol KI 1 mol I 4 mol KI = 0.18 mol I formed CuCl (s) + 4KI(s) CuI(s) + 4KCl(s) + I (s) Initial excess 0.873 mol 0 0 0-0.873 mol +0.18 mol +0.18 mol Since there was no I initially there is 0.18 mol in the amount. If we want the grams of I formed we need the molar mass of I.

0.18 mol I 54 g I 1 mol I = 55.5 g I formed 3. List the general steps required to solve any problem in which you are given the mass of each reactant and asked to calculate the mass of one or more products formed as the result of a complete reaction. 1. Balance the chemical equation which describes the chemical reaction and setup the ICE Table.. Use the molar mass of each substance to convert the grams of the substance to moles. 3. Enter the initial amounts into the ICE Table. 4. Find the entry (a reactant or product) for the row. 5. Use the unit conversion (mole ratio) from the balanced chemical equation to determine the number of moles of the other substances (reactants and products) in the row. Conclude which reactant is in excess and which is limiting. 5. Use the molar mass to convert from moles to grams. 4. In the formation reaction Calculate the number of moles of SO 3 formed when a).0 moles of SO are reacted with 5.0 moles of O Initial.0 mol 5.00 mol 0 But this questions does not tell us which reactant is in excess, so we must determine that ourselve. So we ll assume that all of SO reacts and see what happens in the row of the ICE Table. Initial.0 mol 5.00 mol 0 -.0 mol Assuming that.00 mol SO reacts we will use the stoichiometric coefficients to determine how much O reacts;.0 mol SO 1 mol O mol SO = 1.00 mol O reacts Initial.0 mol 5.00 mol 0

-.0 mol -1.00 mol Since 1.00 mol O reacts and there are 5.00 mol O initially O is in excess and SO is the limiting reagent. We can now calculate how much SO 3 is formed and finish out the ICE table..0 mol SO mol SO 3 mol SO =.00 mol SO 3 formed Initial.0 mol 5.00 mol 0 -.0 mol -1.00 mol.00 mol 0 mol 4.00 mol.00 mol b) 6.0 moles of O are reacted with 4.0 moles SO Initial 4.0 mol 6.00 mol 0 But this question does not tell us which reactant is in excess, so we must determine that ourselve. So we ll assume that all of SO reacts and see what happens in the row of the ICE Table. Initial 4.0 mol 6.00 mol 0-4.0 mol Assuming that 4.00 mol SO reacts we will use the stoichiometric coefficients to determine how much O reacts; 4.0 mol SO 1 mol O mol SO =.00 mol O reacts Initial 4.0 mol 6.00 mol 0-4.0 mol -.00 mol Since.00 mol O reacts and there are 6.00 mol O initially O is in excess and SO is the limiting reagent. We can now calculate how much SO 3 is formed and finish out the ICE table.

4.0 mol SO mol SO 3 mol SO = 4.00 mol SO 3 formed Initial 4.0 mol 6.00 mol 0-4.0 mol -.00 mol 4.00 mol 0 mol 4.00 mol 4.00 mol c) 9.0 moles of O are reacted with 5.0 moles of SO Initial 5.0 mol 9.00 mol 0 But this questions does not tell us which reactant is in excess, so we must determine that ourselve. So we ll assume that all of SO reacts and see what happens in the row of the ICE Table. Initial 5.0 mol 9.00 mol 0-5.0 mol Assuming that 5.00 mol SO reacts we will use the stoichiometric coefficients to determine how much O reacts; 5.0 mol SO 1 mol O mol SO =.50 mol O reacts Initial 5.0 mol 9.00 mol 0-5.0 mol -.50 mol Since.50 mol O reacts and there are 9.00 mol O initially O is in excess and SO is the limiting reagent. We can now calculate how much SO 3 is formed and finish out the ICE table. 5.0 mol SO mol SO 3 mol SO = 5.00 mol SO 3 formed Initial 5.0 mol 9.00 mol 0-5.0 mol -.50 mol 5.00 mol

0 mol 6.50 mol 5.00 mol d) 0.081 moles of SO react with 0.15 moles of O Initial 0.081 mol 0.15 mol 0 But this questions does not tell us which reactant is in excess, so we must determine that ourselve. So we ll assume that all of SO reacts and see what happens in the row of the ICE Table. Initial 0.081 mol 0.15 mol 0-0.081 mol Assuming that 0.081 mol SO reacts we will use the stoichiometric coefficients to determine how much O reacts; 0.081 mol SO 1 mol O mol SO = 0.0406 mol O reacts Initial 0.081 mol 0.15 mol 0-0.081 mol -0.0406 mol Since 0.0406 mol O reacts and there are 0.15 mol O initially O is in excess and SO is the limiting reagent. We can now calculate how much SO 3 is formed and finish out the ICE table. 0.081 mol SO mol SO 3 mol SO = 0.081 mol SO 3 formed Initial 0.081 mol 0.15 mol 0-0.081 mol -0.0406 mol 0.081 mol 0 mol 0.0844 mol 0.081 mol e) 0.0 g SO react with 15.0 g of O

The amounts provided are in grams, so we need to convert those to moles before inserting into the ICE table; 0.0 g SO 1 mol SO 64.0 g = 0.313 mol SO reacting 15.0 g O 1 mol O 3.0 g = 0.469 mol O reacting Initial 0.313 mol 0.469 mol 0 But this questions does not tell us which reactant is in excess, so we must determine that ourselves. So we ll assume that all of SO reacts and see what happens in the row of the ICE Table. Initial 0.313 mol 0.469 mol 0-0.313 mol Assuming that 0.313 mol SO reacts we will use the stoichiometric coefficients to determine how much O reacts; 0.313 mol SO 1 mol O mol SO = 0.157 mol O reacts Initial 0.313 mol 0.469 mol 0-0.313 mol -0.157 mol Since 0.157 mol O reacts and there are 0.469 mol O initially O is in excess and SO is the limiting reagent. We can now calculate how much SO 3 is formed and finish out the ICE table. 0.313 mol SO mol SO 3 mol SO = 0.313 mol SO 3 formed Initial 0.313 mol 0.469 mol 0-0.313 mol -0.157 mol 0 mol 0.31 mol 0.313 mol Now we ll convert the moles of SO 3 to grams

0.313 mol SO 3 80.0 g SO 3 1 mol SO 3 = 5.0 g SO 3 formed

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