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Limiting Reactants Adjusting to Reality Unit 8 WS 3 1 N2 + 3 H2 BEFORE 2 NH3 AFTER How many NH3 molecules are produced? 8 Which reactant is in excess & why? N2 It’s left over How many excess molecules are there? 2 Constructing a BCA Table 1 N2 Equation Before 6 mol + 3 H2 12 mol 2 NH3 0 mol Change After Let’s say you don’t see right away which reactant will get used up…So we’ll just take a guess that the Nitrogen gets used up. Constructing a BCA Table 1 N2 Equation + 3 H2 Before 6 mol 12 mol Change - 6 mol - 18 mol 0 mol WRONG After 2 NH3 0 mol Let’s say you don’t see right away which reactant will get used up…So we’ll just take a guess that the Nitrogen gets used up. Constructing a BCA Table 1 N2 Equation + 3 H2 2 NH3 Before 6 mol 12 mol 0 mol Change - 4 mol -12 mol + 8 mol 2 mol 0 mol 8 mol After Now we know the hydrogen has to get used up. Constructing a BCA Table 1 N2 Equation + 3 H2 2 NH3 Before 6 mol 12 mol 0 mol Change - 4 mol -12 mol + 8 mol 2 mol 0 mol 8 mol After How many NH3 can be produced? 8 Which reactant is in excess & why? N2 There’s some in the after line How many molecules of N2 are there left? 2 What must you look for in a particular reactant mixture to decide what reactant will be in excess? Both the coefficients in the balanced equation & how much of each reactant you start with. # 3. When 0.50 mol of Al reacts with 0.72 mol of I2, how many moles of the excess reactant will remain? How many moles of aluminum iodide will be formed? Equation Before Change After 2 Al + 3 I2 2 Al I3 # 3. When 0.50 mol of Al reacts with 0.72 mol of I2, how many moles of the excess reactant will remain? How many moles of aluminum iodide will be formed? Equation Before Change After 2 Al .50 mol + 3 I2 .72 mol 2 Al I3 # 3. When 0.50 mol of Al reacts with 0.72 mol of I2, how many moles of the excess reactant will remain? How many moles of aluminum iodide will be formed? Equation 2 Al + 3 I2 2 Al I3 Before .50 mol .72 mol 0 mol Change -.48 mol -.72 mol +.48 mol After .02 mol 0 mol .48 mol Looking at these numbers, some of us might see that if there’s a 2 in front of the Al and I have .50 moles of it, I would need .75 moles of the I2 to react with it, & since I have less than that, I know it’s the I2 that runs out. If you don’t see this – you can always guess & test. # 3. When 0.50 mol of Al reacts with 0.72 mol of I2, how many moles of the excess reactant will remain? How many moles of aluminum iodide will be formed? Equation 2 Al + 3 I2 2 Al I3 Before .50 mol .72 mol 0 mol Change -.48 mol -.72 mol +.48 mol After .02 mol 0 mol .48 mol Moles of excess reactant: 0.02 mol Al Moles of AlI3 formed: 0.48 mol AlI3 # 4. When sodium hydroxide reacts with sulfuric acid (H2SO4), water and sodium sulfate are the products. Calculate the mass of sodium sulfate produced when 15.5 g of sodium hydroxide are reacted with 46.7 g of sulfuric acid. Equation Before Change After 2 NaOH + 1 H2SO4 2 H2O + Na2SO4 # 4. When sodium hydroxide reacts with sulfuric acid (H2SO4), water and sodium sulfate are the products. Calculate the mass of sodium sulfate produced when 15.5 g of sodium hydroxide are reacted with 46.7 g of sulfuric acid. 1 mol NaOH 15.5 g NaOH x = 0.388 mol NaOH 40.00 g NaOH 1 mol H2SO4 46.7 g H2SO4 x = 0.473 mol H2SO4 98.72 g H2SO4 # 4. When sodium hydroxide reacts with sulfuric acid (H2SO4), water and sodium sulfate are the products. Calculate the mass of sodium sulfate produced when 15.5 g of sodium hydroxide are reacted with 46.7 g of sulfuric acid. Equation Before Change After 2 NaOH + .388 mol 1 H2SO4 .473 mol 2 H2O 0 mol + Na2SO4 0 mol # 4. When sodium hydroxide reacts with sulfuric acid (H2SO4), water and sodium sulfate are the products. Calculate the mass of sodium sulfate produced when 15.5 g of sodium hydroxide are reacted with 46.7 g of sulfuric acid. Equation 2 NaOH + 1 H2SO4 2 H2O + Na2SO4 Before .388 mol .473 mol 0 mol 0 mol Change -.388 mol -.194 mol +.388 mol +.194 mol 0 mol .279 mol .388 mol .194 mol After Looking at these numbers, some of us might see that if there’s a 2 in front of the NaOH and I have .388 moles of it, I would need only half of that (about 1.9 off the top of my head) for the sulfuric acid, so I must use up the NaOH. If you don’t see this – you can always guess & test. # 4. Calculate the mass of sodium sulfate produced when 15.5 g of sodium hydroxide are reacted with 46.7 g of sulfuric acid. 142.1 g Na2SO4 0.194 mol Na2SO4 x = 27.6 g Na2SO4 1 mol Na2SO4 # 5. A 14.6 g sample of oxygen gas is placed in a sealed container with 2.5 g of hydrogen gas. The mixture is sparked, producing water vapor. Calculate the mass of water formed. Calculate the number of moles of the excess reactant remaining. Equation Before Change After 2 H2 + 1 O2 2 H2O # 5. A 14.6 g sample of oxygen gas is placed in a sealed container with 2.5 g of hydrogen gas. The mixture is sparked, producing water vapor. Calculate the mass of water formed. Calculate the number of moles of the excess reactant remaining. 1 mol O2 14.6 g O2 x = 0.456 mol O2 32.0 g O2 1 mol H2 2.5 g H2 x = 1.24 mol H2 2.02 g H2 # 5. A 14.6 g sample of oxygen gas is placed in a sealed container with 2.5 g of hydrogen gas. The mixture is sparked, producing water vapor. Calculate the mass of water formed. Calculate the number of moles of the excess reactant remaining. Equation Before Change After 2 H2 1.24 mol + 1 O2 0.456 mol 2 H2O 0 mol # 5. A 14.6 g sample of oxygen gas is placed in a sealed container with 2.5 g of hydrogen gas. The mixture is sparked, producing water vapor. Calculate the mass of water formed. Calculate the number of moles of the excess reactant remaining. Equation 2 H2 + 1 O2 2 H2O Before 1.24 mol 0.456 mol 0 mol Change -.912 mol - .456 mol + .912 mol .33 mol 0 mol .912 mol After Looking at these numbers, some of us might see that if there’s a 1 in front of the O2 and I have .456 moles of it, I would need double that (about 0.9 off the top of my head) for the H2, so I must use up the O2. If you don’t see this – you can always guess & test. # 5. Calculate the mass of water formed. Calculate the number of moles of the excess reactant remaining. 18.02 g H2O 0.912 mol H2O x = 16.4 g H2O 1 mol H2O # 6. How much PEA can be made from 75.0g of ammonium formate and 125g of acetophenone? What mass of the excess reactant remains? Equation Before Change After CH5NO2 + C8H8O C8H11N + CO2 + H2O # 6. How much PEA can be made from 75.0g of ammonium formate and 125g of acetophenone? What mass of the excess reactant remains? 1 mol AF 75.0 g AF x = 1.19 mol AF 63.07 g AF 1 mol ACE 125 g ACE x = 1.04 mol ACE 120.16 g ACE # 6. How much PEA can be made from 75.0g of ammonium formate and 125g of acetophenone? What mass of the excess reactant remains? Equation Before Change After CH5NO2 1.19 mol + C8H8O 1.04 mol C8H11N 0 + CO2 0 + H2O 0 # 6. How much PEA can be made from 75.0g of ammonium formate and 125g of acetophenone? What mass of the excess reactant remains? Equation CH5NO2 + C8H8O C8H11N + CO2 + H2O Before 1.19 mol 1.04 mol 0 0 0 Change -1.04 mol -1.04 mol +1.04 mol +1.04 mol +1.04 mol After 0.15 mol 0 mol 1.04 mol 1.04 mol 1.04 mol Looking at these numbers they all have coefficients of 1, so it’s easy to see that the acetophenone is the one that runs out. # 6. How much PEA can be made from 75.0g of ammonium formate and 125g of acetophenone? What mass of the excess reactant remains? 63.07 g AF 0.15 mol AF x = 9.46 g AF in excess 1 mol AF 121.20 g PEA 1.04 mol PEA x = 126 g PEA formed 1 mol PEA