Excess Reactant - teachnlearnchem.com

Excess Reactant - teachnlearnchem.com

Visualizing Limiting Reactant 2 H2 + O2 7 0 8 3 4 2 6 5 mole H2 1 ___ 2.5 1.5 4.5 3.5 1 5 3 mole O2 2 4 ___ Limiting Reactant Excess 2 H 2O 1 5 8 0 4 6 2 3 mole H2O

7 ___ Limiting reactant determines amount of product. Excess Reactant 2 Na + 50 g / 23 g/mol Have 2.17 mol Cl2 / 71 g/mol 0.70 mol LIMITING 2 NaCl 81.9xggNaCl 50 g Need 1.40 mol EXCESS 1:2 coefficients x 58.5 g/mol 1.40 mol Excess Reactant (continued) excess 2 Na

limiting + Cl2 50 g 50 g 2 NaCl 81.9xggNaCl All the chlorine is used up 81.9 g NaCl -50.0 g Cl2 31.9 g Na is consumed in reaction. How much Na is unreacted? 50.0 g - 31.9 g = 18.1 g Na total used excess Conservation of Mass is Obeyed 2 Na + 50 g 2 Na Cl2 81.9xggNaCl 50 g +

50 g Cl2 50 g 2 NaCl 2 NaCl 81.9xggNaCl + Na 18.1 g 31.9 g + 18.1 g 100 g product 81.9 100 g reactant 81.9 Solid aluminum react with chlorine gas to yield solid aluminum chloride. 2 Al(s) excess 125 g + 3 Cl2(g) 2 AlCl3(s) xg excess 125 g If 125 g aluminum react with excess chlorine, how many grams of aluminum chloride are made? x g AlCl3 = 125 g Al Al

1 mol Al 2 mol AlCl3 133.5 g AlCl3 = 618 g AlCl3 2 mol Al 1 mol AlCl3 27 g Al AlCl3 If 125 g chlorine react with excess aluminum, how many grams of aluminum chloride are made? x g AlCl3 = 125 g Cl2 Cl2 1 mol Cl2 71 g Cl2 2 mol AlCl3 133.5 g AlCl3 = 157 g AlCl3 3 mol Cl2 1 mol AlCl3 AlCl3 If 125 g aluminum react with 125 g chlorine, how many grams of aluminum chloride are made? 157 g AlCl3 Were out of Cl2 Solid aluminum react with chlorine gas to yield solid aluminum chloride. If 125 g aluminum react with excess chlorine, how many grams of aluminum chloride are made? 2 Al(s) + 125 g 3 Cl2(g) 2 AlCl3(s) 618xggAlCl3 excess / 27 g/mol

4.6 mol Al x 133.5 g/mol 2:2 Step 1 x g AlCl3 = 125 g Al Al AlCl3 4.6 mol AlCl3 Step 2 Step 3 1 mol Al 2 mol AlCl3 133.5 g AlCl3 = 618 g AlCl3 2 mol Al 1 mol AlCl3 27 g Al Solid aluminum react with chlorine gas to yield solid aluminum chloride. If 125 g chlorine react with excess aluminum, how many grams of aluminum chloride are made? 2 Al(s) excess + 3 Cl2(g) 2 AlCl3(s) 157xggAlCl3 125 g / 71 g/mol

1.76 mol Cl2 x 133.5 g/mol 3:2 1.17 mol AlCl3 3 2 = 1.76 mol Al x mol Al 3x = 3.52 x = 1.17 mol Step 1 x g AlCl3 = 125 g Cl2 Cl2 AlCl3 1 mol Cl2 71 g Cl2 Step 2 Step 3 2 mol AlCl3 133.5 g AlCl3 = 157 g AlCl3 3 mol Cl2 1 mol AlCl3 Limiting Reactant Problems 1. According to the balanced chemical equation, how many atoms of silver will be produced from combining 100 g of copper with 200 g of silver nitrate? Easy Cu(s) + 2 AgNO3(aq) Cu(NO3)2(aq) + 2 Ag(s) . At STP, what volume of laughing gas (dinitrogen monoxide) will be produced from 50 g of nitrogen gas

and 75 g of oxygen gas? Easy . Carbon monoxide can be combined with hydrogen to produce methanol, CH 3OH. Methanol is used as an industrial solvent, as a reactant in some synthesis reactions, and as a clean-burning fuel for some racing cars. If you had 152 kg of carbon monoxide and 24.5 kg of hydrogen gas, how many kilograms Easyof methanol could be produced? . How many grams of water will be produced from 50 g of hydrogen and 100 g of oxygen? Easy Answers: 1. 7.1 x 1023 atoms Ag 2. 40 dm3 N2O 3. 174.3 kg CH3OH 4. 112.5 g H2O Limiting Reactant Problems 1. According to the balanced chemical equation, how many atoms of silver will be produced from combining 100 g of copper with 200 g of silver nitrate? Back Cu(s) + 2 AgNO3(aq) Excess 100 g Limiting 200 g 1.57 x atoms / 170 g/mol / 63.5 g/mol 1.57 mol Cu 1

Cu(NO3)2(aq) + 2 Ag(s) 1.18 mol AgNO3 2 0.59 smaller number is limiting reactant x atoms Ag = 1.18 mol AgNO3 2 mol Ag 6.02 x 1023 atoms Ag 2 mol AgNO3 1 mol Ag = 7.1 x 1023 atoms Ag Limiting Reactant Problems . At STP, what volume of laughing gas (dinitrogen monoxide) will be produced from 50 g of nitrogen gas and 75 g of oxygen gas? Back 2 N2(g) + O2(g) 2 N2O(g) Limiting 50 g Excess 75 g xL / 32 g/mol

/ 28 g/mol 1.79 mol N2 2 2.34 mol O2 1 0.89 2.34 smaller number is limiting reactant x L N2O = 1.79 mol N2 2 mol N2O 22.4 L N2O 2 mol N2 1 mol N2O = 40 L N2O Limiting Reactant Problems . Carbon monoxide can be combined with hydrogen to produce methanol, CH 3OH. Methanol is used as an industrial solvent, as a reactant in some synthesis reactions, and as a clean-burning fuel for some racing cars. If you had 152 kg of carbon monoxide and 24.5 kg of hydrogen gas, how many kilograms of methanol could be produced? Back CO (g) + 2 H2(g) CH3OH (g) Limiting 152.5 g Excess 24.5 g

x g kg 174.3 / 2 g/mol / 28 g/mol 5.45 mol CO 1 12.25 mol H2 2 5.45 6.125 smaller number is limiting reactant x g CH3OH = 5.45 mol CO Work the entire problem with the mass in grams. At the end, change answer to units of kilograms. 1 mol CH3OH 1 mol CO 32 g CH3OH 1 mol CH3OH = 174.3 g CH3OH Limiting Reactant Problems . How many grams of water will be produced from 50 g of hydrogen and 100 g of oxygen? Back 2 H2(g) + Excess

50 g O2(g) 2 H2O(g) Limiting 100 g xg / 32 g/mol / 2 g/mol 25 mol H2 2 3.125 mol O2 1 12.5 3.125 smaller number is limiting reactant x g H2O = 3.125 mol O2 2 mol H2O 18 g H2O 1 mol O2 1 mol H2O = 112.5 g H2O Limiting Reactant Problems - continued 5. An unbalanced chemical equation is given as __N 2H24(l) + __N2O4(l) If you begin with 400 g of N2H4 and 900 g of N2O4 __N2(g) 3 + __ H2O(g).

4 Easy A. Find the number of liters of water produced (at STP), assuming the reaction goes to completion. Easy B. Find the number of liters of nitrogen produced at STP, assuming the reaction goes to completion. Easy C. Find the mass of excess reactant left over at the conclusion of the reaction. 4 + __O 2(g) . An unbalanced chemical equation is given as __Na(s) If you have 100 g of sodium and 60 g of oxygen __Na22 O (s) Easy A. Find the number of moles of sodium oxide produced. Easy B. Find the mass of excess reactant left over at the conclusion of the reaction. Answers: 5A. 560 L H2O (@STP - gas) or 0.45 L H2O 5B. 420 L N2 5C. 325 g N2O4 excess 6A. 2.17 mol Na2O 6B. 25.2 g O2 excess Limiting Reactant Problems

5. An unbalanced chemical equation is given as __N 2H24(l) + __N2O4(l) If you begin with 400 g of N2H4 and 900 g of N2O4 __N2(g) 3 + __ H2O(g). 4 A. Find the number of liters of water produced at STP, assuming the reaction goes to completion. Back 2 N2H4(l) + N2O4(l) 400 g 6.25 xL 900 g / 92 g/mol / 32 g/mol 12.5 mol N2H4 2 3 N2(g) + 4 H2O(g) 9.78 mol N2O4 1 9.78 smaller number is limiting reactant Water Density is a SOLID of water at STP this is isnt 1.0possible! g/mL

4 mol H2O 22.4 L H2O 4 mol O 18 g H2O 1 mL H2O = 560 1 L LHH x L H2O = 12.5 mol N2HH 2O 2O 4 2 = 0.45 L H2O x L H2O = 12.5 mol N2H4 2 mol N2H4 1 mol H2O 2 mol N2H4 1 mol H2O 1.0 g H2O 1000 mL H2O Limiting Reactant Problems 5. An unbalanced chemical equation is given as __N 2H24(l) + __N2O4(l) If you begin with 400 g of N2H4 and 900 g of N2O4 __N2(g) 3 + __ H2O(l). 4 A. Find the number of liters of water produced, assuming the reaction goes to completion. Back 2 N2H4(l) + N2O4(l) 400 g xL 900 g / 92 g/mol / 32 g/mol 12.5 mol N2H4 2 3 N2(g) + 4 H2O(g) 9.78 mol N2O4

1 6.25 Density of water is 1.0 g/mL 9.78 smaller number is limiting reactant x L H2O = 12.5 mol N2H4 4 mol H2O 18 g H2O 1 mL H2O 1 L H2O 2 mol N2H4 1 mol H2O 1.0 g H2O 1000 mL H2O = 0.45 L H2O Limiting Reactant Problems 5. An unbalanced chemical equation is given as __N 2H24(l) + __N2O4(l) If you begin with 400 g of N2H4 and 900 g of N2O4 __N2(g) 3 + __ H2O(g). 4 B. Find the number of liters of nitrogen produced at STP, assuming the reaction goes to completion. Back 2 N2H4(l) + N2O4(l) 400 g 6.25 xL

900 g / 92 g/mol / 32 g/mol 12.5 mol N2H4 2 3 N2(g) + 4 H2O(g) 9.78 mol N2O4 1 9.78 smaller number is limiting reactant x L N2 = 12.5 mol N2H4 3 mol N2 22.4 L N2 2 mol N2H4 1 mol N2 = 420 L N2 Limiting Reactant Problems 5. An unbalanced chemical equation is given as __N 2H24(l) + __N2O4(l) If you begin with 400 g of N2H4 and 900 g of N2O4 __N2(g) 3 + __ H2O(g). 4 C. Find the mass of excess reactant left over at the conclusion of the reaction. Back 2 N2H4(l) + N2O4(l) 400 g 575

x gg / 32 g/mol 12.5 mol N2H4 N2(g) + H2O(g) x 92 g/mol 900 g N2O4 have needed 325 g N2O4 excess 6.25 mol N2O4 x g N2O4 = 12.5 mol N2H4 1 mol N2O4 92 g N2O4 2 mol N2H4 1 mol N2O4 = 575 g N2O4 Limiting Reactant Problems . An unbalanced chemical equation is given as __Na(s) 4 + __O 2(g) If you have 100 g of sodium and 60 g of oxygen Back __Na22O (s) A. Find the number of moles of sodium oxide produced. 4 Na(s) + O2 (g)

100 g 2 Na2O (s) x mol 60 g / 32 g/mol / 23 g/mol 4.35 mol Na 4 1.875 mol O2 1 1.087 1.875 smaller number is limiting reactant x mol Na2O = 4.35 mol Na 2 mol Na2O 4 mol Na = 2.17 mol Na2O Limiting Reactant Problems . An unbalanced chemical equation is given as __Na(s) 4 + __O 2(g) If you have 100 g of sodium and 60 g of oxygen __Na2O 2 (s) B. Find the mass of excess reactant left over at the conclusion of the reaction. Back 4 Na(s)

+ 100 g O2 (g) 2 Na2O (s) 34.8 x gg / 23 g/mol x 32 g/mol - 60 g O2 25.2 g O2 4.35 mol Na have needed excess 1.087 mol O2 x g O2 = 4.35 mol Na 1 mol O2 32 g O2 4 mol Na 1 mol O2 = 34.8 g O2 Percent Yield measured in lab % yield =

actual yield theoretical yield calculated on paper x 100 When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. actual yield 46.3 g K2CO3 + 2HCl 45.8 g excess 2KCl + H2O CO+3 CO2 ?g theoretical yield Theoretical yield x g KCl = 45.8 g K2CO3 % Yield = 1 mol K2CO3 2 mol KCl 74.5 g KCl = 49.4 g KCl 138 g K2CO3 1 mol K2CO3 1 mol KCl Actual Yield Theoretical Yield % Yield = 46.3 g KCl x 100

% Yield = 93.7% efficient Percent Yield actual 500yield g 0.80 % 80 yield == x g yield theoretical Need 500 g of Y % yield = 80% x 100 % actual yield W2 + 2X x atoms Y 0.80 x = 500 g 0.80 0.80 x = 625 g xg x L @STP theoretical yield x atoms W = 625 g Y x L X = 625 g Y 1 mol Y 1 mol W2 89 g Y 1 mol Y

6.02 x 1023 molecules W2 1 mol W2 1 mol Y 2 mol X 22.4 L X 89 g Y 1 mol Y 1 mol X 2 atoms W 1 molecule W2 x 1024 atoms W W 315 LLX8.45 Xx 10 = 315 8.45 atoms 24 Cartoon courtesy of NearingZero.net Print Print Copy Copy ofofLab Lab Baking Soda Lab Na NaHCO HCl

HCO3 + H Cl sodium bicarbonate + H2CO3 hydrochloric acid sodium chloride baking soda table salt (l) + H2O (g) heat CO2 (g) actual yield ?g NaHCO3 + HCl 5g excess actual yield theoretical yield gas NaCl + H2O + CO2 xg theoretical yield

% yield = gas x 100 % Resources - Stoichiometry Objectives - stoichiometry Objectives - mole / chemical formula Lab nuts and bolts Lab - baking soda lab Episode 11 The Mole Smores activity Worksheet - careers in chemistry: farming key Worksheet - careers in chemistry: dentistry key Worksheet - easy stoichiometry Worksheet - energy Worksheet - generic Worksheet moles and mass relationships Worksheet - percent yield Worksheet - stoichiometry problems 1 Worksheet - limiting reactants Worksheet - vocabulary Worksheet - lecture outline Textbook - questions Worksheet visualizing limiting reactant Outline (general)

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